An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V1) Cambridge International A Level Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
Paper 11 Multiple Choice (Core)
There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct.
41 Question · 41 marks
Question 1 · multiple-choice
1 marks
A student prepares a sample of hydrated copper(II) sulfate crystals by reacting excess copper(II) oxide with dilute sulfuric acid.
Why is excess copper(II) oxide added?
A.to make sure all the sulfuric acid has reacted
B.to increase the rate of crystallization
C.to act as a catalyst in the reaction
D.to prevent the copper(II) sulfate from dissolving
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Worked solution
Excess copper(II) oxide (an insoluble base) is added to ensure that all of the dilute sulfuric acid is completely reacted and neutralized. This ensures the resulting copper(II) sulfate solution is not contaminated with unreacted acid. The unreacted excess copper(II) oxide can then be easily removed by filtration.
Marking scheme
Award 1 mark for the correct option A. - Reject B: Excess reactant does not increase the rate of crystallization. - Reject C: Copper(II) oxide is a reactant, not a catalyst. - Reject D: Excess solid does not reduce the amount of water to be evaporated.
Question 2 · multiple-choice
1 marks
Molten lead(II) bromide is electrolysed using inert carbon electrodes.
Which row correctly describes the products formed at the electrodes?
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Worked solution
During the electrolysis of molten lead(II) bromide (\(\text{PbBr}_2\)), the lead(II) ions (\(\text{Pb}^{2+}\)) migrate to the negative electrode (cathode) where they gain electrons to form lead metal. The bromide ions (\(\text{Br}^-\)) migrate to the positive electrode (anode) where they lose electrons to form bromine gas. Therefore, bromine is formed at the anode and lead is formed at the cathode.
Marking scheme
Award 1 mark for the correct option A. - Reject B: The products are reversed. - Reject C and D: Hydrogen is not produced because there is no water present in a molten electrolyte.
Question 3 · multiple-choice
1 marks
Which statement about addition polymerisation is correct?
A.The monomer molecules must contain a carbon-carbon double bond.
B.A small molecule, such as water, is also produced during the reaction.
C.The polymer has the same physical properties as the monomer.
D.The reaction requires a metal catalyst to form ionic bonds.
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Worked solution
In addition polymerisation, unsaturated monomer molecules containing carbon-carbon double bonds (\(\text{C=C}\)), such as ethene, react together to form a polymer chain containing only single bonds. No other product is formed during this process.
Marking scheme
Award 1 mark for the correct option A. - Reject B: Condensation polymerisation, not addition polymerisation, produces a small molecule like water. - Reject C: Polymers have very different physical properties compared to their monomers (e.g., state, melting point). - Reject D: Polymerisation of alkenes involves covalent bonding, not ionic bonding.
Question 4 · multiple-choice
1 marks
Molten lead(II) bromide is electrolysed using inert carbon electrodes. Which row correctly describes the products and observations at each electrode?
A.Cathode product: lead, Cathode observation: grey liquid; Anode product: bromine, Anode observation: brown gas
D.Cathode product: hydrogen, Cathode observation: colourless gas; Anode product: bromine, Anode observation: brown gas
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Worked solution
During the electrolysis of molten lead(II) bromide, positive lead ions (\(\text{Pb}^{2+}\)) migrate to the negative electrode (cathode) where they gain electrons to form lead metal. Since the process is carried out at high temperature, lead is formed as a grey liquid. Negative bromide ions (\(\text{Br}^-\)) migrate to the positive electrode (anode) where they lose electrons to form bromine gas, which is observed as a brown gas. Thus, row A is correct.
Marking scheme
1 mark for the correct answer A. 0 marks for other answers.
Question 5 · multiple-choice
1 marks
Which sequence of experimental steps is used to prepare a pure, dry sample of hydrated copper(II) sulfate crystals from insoluble copper(II) oxide and dilute sulfuric acid?
A.Add excess copper(II) oxide to dilute sulfuric acid, filter the mixture, heat the filtrate to the crystallisation point, allow it to cool, and filter the crystals.
B.Add excess copper(II) oxide to dilute sulfuric acid, heat the mixture to dryness, and wash the remaining solid with hot water.
C.Add excess dilute sulfuric acid to copper(II) oxide, filter the mixture to collect the solid residue, and dry the solid in an oven.
D.Add equal volumes of dilute sulfuric acid and water to copper(II) oxide, filter, and evaporate all the water to leave anhydrous crystals.
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Worked solution
To prepare a soluble salt from an insoluble oxide: 1. Add excess copper(II) oxide to the dilute sulfuric acid to ensure all the acid is neutralised. 2. Filter the mixture to remove the excess unreacted solid. 3. Heat the filtrate (copper(II) sulfate solution) to the crystallisation point (do not evaporate to dryness, as this would form anhydrous powder instead of hydrated crystals). 4. Allow the hot saturated solution to cool so that crystals can form. 5. Filter the crystals to separate them from the remaining solution, and then dry them.
Marking scheme
1 mark for option A. 0 marks for other options.
Question 6 · multiple-choice
1 marks
Ethene is a monomer that can be polymerised to form poly(ethene). Which statement about this polymerisation and the polymer formed is correct?
A.Ethene molecules join together by condensation polymerisation to form poly(ethene).
B.Ethene contains only carbon-carbon single bonds which break during the reaction.
C.Many ethene molecules join together to form a long-chain polymer called poly(ethene).
D.Poly(ethene) is a biodegradable plastic that is easily broken down by bacteria in the soil.
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Worked solution
Ethene undergoes addition polymerisation (not condensation polymerisation) because the double bond in ethene opens up to join many monomer molecules together without forming any other products. Therefore, many ethene molecules join to form a long-chain polymer called poly(ethene). Poly(ethene) is non-biodegradable, meaning it is not easily broken down by bacteria.
Marking scheme
1 mark for option C. 0 marks for other options.
Question 7 · multiple-choice
1 marks
A student performs a paper chromatography experiment to analyze a black ink. The starting line is drawn in pencil and a spot of ink is placed on it. After running the chromatogram, the solvent front has moved \(8.0\text{ cm}\) from the starting line. The ink separates into two spots. The distance from the starting line to the center of the faster-moving spot is \(6.0\text{ cm}\). What is the \(R_f\) value of this spot?
A.0.25
B.0.75
C.1.33
D.6.02
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Worked solution
The formula for the Retention Factor is: \(R_f = \frac{\text{distance traveled by substance}}{\text{distance traveled by solvent front}}\). For the faster-moving spot: \(R_f = \frac{6.0\text{ cm}}{8.0\text{ cm}} = 0.75\).
Marking scheme
1 mark for the correct calculation: 6.0 / 8.0 = 0.75.
Question 8 · multiple-choice
1 marks
Dilute hydrochloric acid is added to a solid substance in a test-tube. Rapid effervescence is observed, and the gas produced turns limewater cloudy. Which substance was added to the acid?
A.copper
B.sodium hydroxide
C.calcium carbonate
D.zinc oxide
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Worked solution
Acids react with metal carbonates to produce salt, water, and carbon dioxide gas. Carbon dioxide gas turns limewater cloudy. Calcium carbonate reacts with dilute hydrochloric acid to produce calcium chloride, water, and carbon dioxide gas: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\). None of the other options react to produce carbon dioxide (copper does not react, sodium hydroxide forms water and a soluble salt, and zinc oxide forms water and a soluble salt).
Marking scheme
1 mark for identifying calcium carbonate as the solid that produces carbon dioxide gas on reaction with acid.
Question 9 · multiple-choice
1 marks
Which statement describes the trends in the physical state at room temperature and the reactivity of the Group VII elements (halogens) as the group is descended (from fluorine to iodine)?
A.The physical states change from gas to solid, and reactivity increases.
B.The physical states change from solid to gas, and reactivity decreases.
C.The physical states change from gas to solid, and reactivity decreases.
D.The physical states change from solid to gas, and reactivity increases.
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Worked solution
As Group VII is descended (from fluorine to iodine), the physical states change from gas (fluorine and chlorine are gases) to liquid (bromine is a liquid) to solid (iodine is a solid). The reactivity of the halogens decreases down the group because the atomic radius increases, making it harder for the atom to attract and gain an electron.
Marking scheme
1 mark for correctly identifying that states change from gas to solid and reactivity decreases down the group.
Question 10 · multiple-choice
1 marks
A student has a mixture of sand and solid copper(II) sulfate. The student wants to obtain a pure, dry sample of copper(II) sulfate crystals. Which sequence of steps should the student use?
A.Add water and stir \rightarrow filter \rightarrow heat the filtrate to crystallising point, then leave to cool \rightarrow dry the crystals
B.Add water and stir \rightarrow heat to dryness \rightarrow filter \rightarrow dry the crystals
C.Filter \rightarrow add water and stir \rightarrow heat to dryness \rightarrow dry the crystals
D.Heat the mixture \rightarrow add water and stir \rightarrow filter \rightarrow dry the crystals
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Worked solution
To separate an insoluble solid (sand) from a soluble salt (copper(II) sulfate): 1. Add water and stir so that the copper(II) sulfate dissolves while the sand remains insoluble. 2. Filter the mixture to remove the insoluble sand as residue, leaving the copper(II) sulfate solution as the filtrate. 3. Heat the filtrate to the crystallising point (until a saturated solution is formed) and leave it to cool so that crystals form. 4. Dry the crystals between pieces of filter paper.
Marking scheme
A is correct (1 mark) because it correctly lists dissolution, filtration, partial evaporation, and drying. B is incorrect because filtering must precede heating, and heating to dryness does not produce crystals. C is incorrect because dry filtering does not separate the solids. D is incorrect because heating the dry mixture is not part of the separation process.
Question 11 · multiple-choice
1 marks
Molten lead(II) bromide is electrolysed using inert carbon electrodes. Which statement about this electrolysis is correct?
A.Lead is formed at the negative electrode (cathode) and bromine is formed at the positive electrode (anode).
B.Lead is formed at the positive electrode (anode) and bromine is formed at the negative electrode (cathode).
C.Hydrogen is formed at the negative electrode (cathode) and oxygen is formed at the positive electrode (anode).
D.Bromine is formed at the negative electrode (cathode) and lead is formed at the positive electrode (anode).
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Worked solution
During the electrolysis of molten lead(II) bromide (\(\text{PbBr}_2\)), the positive lead ions (\(\text{Pb}^{2+}\)) move to the negative electrode (cathode) where they gain electrons to form lead metal. The negative bromide ions (\(\text{Br}^-\)) move to the positive electrode (anode) where they lose electrons to form bromine gas.
Marking scheme
A is correct (1 mark) because lead metal is formed at the cathode and bromine gas is formed at the anode. B and D are incorrect because the products are assigned to the wrong electrodes. C is incorrect because molten lead(II) bromide contains no water, so hydrogen and oxygen are not produced.
Question 12 · multiple-choice
1 marks
Ethene, \(\text{C}_2\text{H}_4\), reacts to form the polymer poly(ethene). Which row describes the type of polymerisation reaction and the change in bonding that occurs?
A.reaction type: addition; bonding change: double bonds become single bonds
B.reaction type: addition; bonding change: single bonds become double bonds
C.reaction type: condensation; bonding change: double bonds become single bonds
D.reaction type: condensation; bonding change: single bonds become double bonds
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Worked solution
Ethene molecules join together to form poly(ethene) in an addition polymerisation reaction. In this process, the carbon-carbon double bonds (\(\text{C}=\text{C}\)) in the ethene monomers open up to become carbon-carbon single bonds (\(\text{C}-\text{C}\)) in the long polymer chain.
Marking scheme
A is correct (1 mark) because poly(ethene) is formed by addition polymerisation and involves double bonds converting to single bonds. B is incorrect because double bonds do not form from single bonds. C and D are incorrect because poly(ethene) is not formed by condensation polymerisation.
Question 13 · multiple-choice
1 marks
A mixture contains insoluble sand, soluble sodium chloride, and water. Which sequence of processes will produce a pure sample of dry sand and a pure sample of dry sodium chloride crystals?
A.Filter the mixture to obtain the sand as residue; wash and dry it. Heat the filtrate to crystallisation point, then leave it to cool and dry the crystals.
B.Filter the mixture to obtain the sand; heat the sand to crystallisation point. Distil the filtrate to obtain dry sodium chloride.
C.Distil the mixture to obtain dry sodium chloride. Filter the residue to obtain the sand, then wash and dry it with alcohol.
D.Evaporate the entire mixture to dryness. Dissolve the remaining solid in alcohol, filter to obtain the sand, then dry the crystals.
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Worked solution
Filtration separates the insoluble sand from the soluble sodium chloride solution. The sand residue is washed with distilled water to remove any remaining salt solution and then dried to obtain pure sand. The filtrate, which is sodium chloride solution, is heated to concentrate it (to the crystallisation point), allowed to cool to form crystals, and the crystals are then dried.
Marking scheme
1 mark for the correct option (A).
Question 14 · multiple-choice
1 marks
Aqueous chlorine is added to a test-tube containing aqueous potassium bromide. What is the observation and the correct explanation for this reaction?
A.The solution turns orange-brown because chlorine is more reactive than bromine and displaces bromide ions.
B.The solution turns purple because chlorine is less reactive than bromine and is displaced by bromide ions.
C.The solution remains colourless because chlorine is less reactive than bromine.
D.The solution turns orange-brown because bromine is more reactive than chlorine and displaces chloride ions.
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Worked solution
Chlorine is higher in Group VII than bromine, meaning chlorine is more reactive than bromine. When aqueous chlorine is added to potassium bromide solution, chlorine displaces the less reactive bromide ions, forming aqueous bromine. Aqueous bromine causes the solution to change from colourless to orange-brown: \(\text{Cl}_2(\text{aq}) + 2\text{KBr}(\text{aq}) \rightarrow 2\text{KCl}(\text{aq}) + \text{Br}_2(\text{aq})\).
Marking scheme
1 mark for the correct option (A).
Question 15 · multiple-choice
1 marks
A student is given two separate tubes: one contains ethane gas and the other contains ethene gas. Which test and observation can be used to identify the tube containing ethene?
A.Add aqueous bromine; the mixture turns from orange to colourless.
B.Add aqueous bromine; the mixture turns from colourless to orange.
C.Introduce a lighted splint; the gas burns with a squeaky pop.
D.Introduce damp blue litmus paper; the paper turns red.
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Worked solution
Ethene is an alkene, which is an unsaturated hydrocarbon containing a carbon-carbon double bond (\(\text{C}=\text{C}\)). Unsaturated hydrocarbons react rapidly with aqueous bromine (bromine water) in an addition reaction, causing the orange bromine water to decolourise (turn colourless). Ethane is a saturated alkane and does not react with bromine water under these conditions.
Marking scheme
1 mark for the correct option (A).
Question 16 · multiple-choice
1 marks
Why is the baseline (start line) on a chromatography paper drawn in pencil rather than in ink?
A.Pencil graphite does not dissolve in the solvent and will not run up the paper.
B.Pencil graphite has a higher Rf value than any component in the ink.
C.The ink would react chemically with the chromatography paper.
D.The ink prevents the solvent from moving up the chromatography paper.
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Worked solution
Pencil lead is made of graphite, which is insoluble in chromatography solvents. If ink were used, it would dissolve in the solvent and move up the paper along with the mixture being tested, interfering with the results. Therefore, option A is the correct answer.
Marking scheme
1 mark for selecting the correct option A.
Question 17 · multiple-choice
1 marks
Which statement describes the trends in the physical properties of the Group VII halogens as the group is descended (from fluorine to iodine)?
A.The colors of the elements become darker and their melting points increase.
B.The colors of the elements become lighter and their melting points increase.
C.The colors of the elements become darker and their melting points decrease.
D.The colors of the elements become lighter and their melting points decrease.
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Worked solution
As you descend Group VII: 1. The colors of the halogens become progressively darker (fluorine is a pale yellow gas, chlorine is a pale green gas, bromine is a red-brown liquid, and iodine is a grey-black solid). 2. Their melting and boiling points increase due to the increase in strength of the intermolecular forces as the molecular size increases.
Marking scheme
1 mark for selecting the correct option A.
Question 18 · multiple-choice
1 marks
A student wants to separate a mixture of solid copper(II) oxide (insoluble in water) and solid sodium chloride (soluble in water) to obtain pure, dry samples of both.
Which sequence of steps should the student use?
A.Add water, stir, filter, wash and dry the residue, and evaporate the filtrate to dryness.
B.Add water, stir, evaporate the mixture to dryness, and then filter.
C.Filter the dry solid mixture, wash the residue with water, and evaporate the filtrate.
D.Evaporate the dry solid mixture, add water, filter, and dry the residue.
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Worked solution
To separate an insoluble solid from a soluble solid: 1. Add water and stir: the soluble sodium chloride dissolves, while the insoluble copper(II) oxide remains as a solid suspension. 2. Filter: the copper(II) oxide remains on the filter paper as the residue. Wash and dry it to get pure, dry copper(II) oxide. 3. Evaporate the filtrate (sodium chloride solution): the water evaporates, leaving behind pure, dry sodium chloride crystals.
Marking scheme
1 mark for selecting the correct option A.
Question 19 · multiple-choice
1 marks
A student prepares a pure, dry sample of copper(II) sulfate crystals. They add excess copper(II) oxide to hot dilute sulfuric acid. Which sequence of steps should the student carry out next to obtain pure, dry crystals of copper(II) sulfate?
A.Filter the mixture, then heat the filtrate until all the water has evaporated.
B.Filter the mixture, heat the filtrate until a saturated solution is formed, allow to cool and crystallize, then filter and dry the crystals.
C.Heat the mixture until all the water has evaporated, then filter the dry solid.
D.Allow the mixture to cool and crystallize, filter to obtain the crystals, then wash them with hot water.
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Worked solution
To obtain pure, dry crystals of copper(II) sulfate from a mixture containing excess insoluble copper(II) oxide: 1. Filter the mixture to remove the unreacted copper(II) oxide residue. 2. Heat the filtrate (copper(II) sulfate solution) to evaporate some water until a saturated solution is formed. 3. Allow the hot saturated solution to cool slowly so that crystals form. 4. Filter the crystals from the remaining solution, then dry them (e.g., between sheets of filter paper).
Marking scheme
Award 1 mark for the correct option B. All other options describe incorrect sequence of steps or inappropriate techniques.
Question 20 · multiple-choice
1 marks
Molten lead(II) bromide is electrolyzed using inert carbon electrodes. What are the observations at the anode (positive electrode) and at the cathode (negative electrode)?
A.Anode: bubbles of colorless gas; Cathode: grey liquid
B.Anode: brown gas; Cathode: grey liquid
C.Anode: grey solid; Cathode: brown gas
D.Anode: bubbles of colorless gas; Cathode: reddish-brown solid
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Worked solution
During the electrolysis of molten lead(II) bromide: - At the anode (positive electrode), bromide ions (\(\text{Br}^-\)) are oxidized to form bromine molecules (\(\text{Br}_2\)), which is observed as a brown gas. - At the cathode (negative electrode), lead ions (\(\text{Pb}^{2+}\)) are reduced to form lead metal, which is observed as a grey liquid at these high temperatures.
Marking scheme
Award 1 mark for the correct option B. Identify that bromine gas is brown and lead is grey.
Question 21 · multiple-choice
1 marks
Which row correctly identifies the method used to obtain pure water from sea water, and ethanol from a mixture of ethanol and water?
A.Pure water from sea water: simple distillation; Ethanol from ethanol and water: fractional distillation
B.Pure water from sea water: filtration; Ethanol from ethanol and water: simple distillation
C.Pure water from sea water: crystallization; Ethanol from ethanol and water: filtration
D.Pure water from sea water: simple distillation; Ethanol from ethanol and water: crystallization
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Worked solution
Simple distillation is used to separate and collect a pure liquid (solvent) from a mixture containing non-volatile dissolved solids (like salt in sea water). Fractional distillation is used to separate miscible liquids (like ethanol and water) that have different boiling points.
Marking scheme
Award 1 mark for the correct option A. Options B, C, and D are incorrect separation techniques for the given mixtures.
Question 22 · multiple-choice
1 marks
A solid mixture contains insoluble barium sulfate and soluble sodium chloride. Which sequence of steps is used to obtain a pure, dry sample of barium sulfate from this mixture?
A.Add water, filter, and then dry the residue on the filter paper.
B.Add water, evaporate the liquid, and then filter.
C.Heat the mixture until it melts, and then filter.
D.Add water, filter, and then evaporate the filtrate to dryness.
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Worked solution
Barium sulfate is insoluble in water, whereas sodium chloride is soluble. When water is added, only the sodium chloride dissolves. Filtering this mixture leaves the insoluble barium sulfate on the filter paper as the residue, which can then be washed and dried. Evaporating the filtrate would instead recover the sodium chloride.
Marking scheme
Award 1 mark for the correct option A. Reject other options because they either do not separate the components (B, C) or isolate the wrong component (D).
Question 23 · multiple-choice
1 marks
An atom of sodium has a nucleon (mass) number of 23 and a proton number of 11. Which option correctly shows the number of protons, neutrons and electrons in a sodium ion, \(\text{Na}^+\)?
A.Protons: 11, Neutrons: 12, Electrons: 10
B.Protons: 11, Neutrons: 12, Electrons: 11
C.Protons: 11, Neutrons: 23, Electrons: 10
D.Protons: 12, Neutrons: 11, Electrons: 11
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Worked solution
The proton number is 11, so there are 11 protons. The number of neutrons is the nucleon number minus the proton number, which is \(23 - 11 = 12\). A neutral sodium atom has 11 electrons, so a sodium ion with a \(1+\) charge (\(\text{Na}^+\)) has lost one electron, leaving it with 10 electrons.
Marking scheme
Award 1 mark for the correct option A. Reject options with incorrect neutron or electron counts.
Question 24 · multiple-choice
1 marks
Molten lead(II) bromide is electrolysed using inert carbon electrodes. Which row correctly describes the products formed at the positive electrode (anode) and the negative electrode (cathode)?
A.Anode: lead, Cathode: bromine
B.Anode: oxygen, Cathode: hydrogen
C.Anode: bromine, Cathode: lead
D.Anode: hydrogen, Cathode: oxygen
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Worked solution
During the electrolysis of molten lead(II) bromide, the positive electrode (anode) attracts negative bromide ions, which are oxidized to form bromine. The negative electrode (cathode) attracts positive lead(II) ions, which are reduced to form lead metal.
Marking scheme
Award 1 mark for the correct option C. Reject options showing incorrect electrode products or reversed electrodes.
Question 25 · MCQ
1 marks
Molten lead(II) bromide is electrolysed using inert carbon electrodes. Which row correctly describes the product and observation at each electrode?
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Worked solution
During the electrolysis of molten lead(II) bromide, positive lead ions, \(\text{Pb}^{2+}\), move to the negative electrode (cathode) where they gain electrons to form metallic lead, which is a grey liquid at this high temperature. Negative bromide ions, \(\text{Br}^{-}\), move to the positive electrode (anode) where they lose electrons to form bromine gas, which is seen as a brown vapour. Therefore, row A is correct.
Marking scheme
1 mark for the correct option A.
Question 26 · MCQ
1 marks
An atom of sodium has a proton number of 11 and a nucleon number of 23. How many protons, neutrons and electrons are in this neutral atom?
A.11 protons, 12 neutrons, 11 electrons
B.11 protons, 11 neutrons, 12 electrons
C.12 protons, 11 neutrons, 11 electrons
D.11 protons, 12 neutrons, 12 electrons
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Worked solution
The proton number is 11, which tells us there are 11 protons in the nucleus. In a neutral atom, the number of electrons is equal to the number of protons, so there are 11 electrons. The nucleon number is 23, which is the sum of protons and neutrons. The number of neutrons is calculated as \(23 - 11 = 12\). Therefore, there are 11 protons, 12 neutrons and 11 electrons.
Marking scheme
1 mark for the correct option A.
Question 27 · MCQ
1 marks
Which statement about the Group VII elements (halogens) is correct?
A.Chlorine is a pale yellow-green gas, and the melting points of the halogens increase down the group.
B.Bromine is a red-brown liquid, and the density of the halogens decreases down the group.
C.Iodine is a grey-black solid, and its reactivity is greater than that of chlorine.
D.All halogens exist as monatomic gases at room temperature.
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Worked solution
Chlorine is a pale yellow-green gas at room temperature. The melting points of the halogens increase as you go down the group (fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid). This makes statement A correct. Statement B is incorrect because density increases down the group. Statement C is incorrect because reactivity decreases down the group. Statement D is incorrect because halogens exist as diatomic molecules, not monatomic.
Marking scheme
1 mark for the correct option A.
Question 28 · multiple-choice
1 marks
A student wants to prepare pure, dry crystals of copper(II) sulfate by reacting excess copper(II) oxide with dilute sulfuric acid. Which sequence of steps should be carried out after the reaction is complete?
A.Filter the mixture to remove excess copper(II) oxide \(\rightarrow\) heat the filtrate to the crystallisation point \(\rightarrow\) cool to crystallise \(\rightarrow\) filter and dry the crystals
B.Evaporate the entire mixture to dryness \(\rightarrow\) cool to crystallise \(\rightarrow\) wash the residue with dilute sulfuric acid
C.Filter the mixture to remove excess copper(II) oxide \(\rightarrow\) evaporate the filtrate to dryness \(\rightarrow\) wash the crystals with hot water
D.Heat the mixture to the crystallisation point \(\rightarrow\) cool to crystallise \(\rightarrow\) filter the mixture to remove excess copper(II) oxide
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Worked solution
To obtain pure, dry crystals of a soluble salt from an insoluble reactant and an acid: 1. Filter the mixture to remove the unreacted excess copper(II) oxide. 2. Heat the filtrate until the crystallisation point is reached (do not evaporate to dryness, as this would produce anhydrous powder instead of hydrated crystals). 3. Allow the hot, saturated solution to cool so that crystals can form. 4. Filter the crystals from the remaining liquid and dry them using filter paper.
Marking scheme
1 mark for the correct sequence of steps: filtration of excess reactant, partial evaporation (crystallisation), cooling, and final filtration/drying.
Question 29 · multiple-choice
1 marks
Molten lead(II) bromide is electrolysed using inert carbon electrodes. Which observations are made at the anode and at the cathode?
A.brown gas at the anode, grey liquid at the cathode
B.grey liquid at the anode, brown gas at the cathode
C.bubbles of a colourless gas at the anode, grey liquid at the cathode
D.brown gas at the anode, bubbles of a colourless gas at the cathode
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Worked solution
During the electrolysis of molten lead(II) bromide, positive lead ions move to the negative electrode (cathode) where they are reduced to metallic lead, which is seen as a shiny grey liquid. Negative bromide ions move to the positive electrode (anode) where they are oxidised to form bromine gas, which is observed as brown fumes.
Marking scheme
1 mark for identifying brown gas at the anode and a grey liquid at the cathode.
Question 30 · multiple-choice
1 marks
Which statement about the polymerisation of ethene to form poly(ethene) is correct?
A.It is an addition reaction where many ethene molecules join together to form a single large molecule.
B.It is a condensation reaction where water is produced as a side-product.
C.The monomer ethene is a saturated hydrocarbon.
D.The polymer poly(ethene) contains many carbon-carbon double bonds.
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Worked solution
Poly(ethene) is formed by addition polymerisation, where thousands of unsaturated ethene molecules (monomers) join together by the opening of their double bonds to form a single, long-chain saturated molecule (the polymer) without forming any side products.
Marking scheme
1 mark for identifying addition polymerisation of ethene molecules into a single large molecule.
Question 31 · multiple_choice
1 marks
Molten lead(II) bromide is electrolysed using inert graphite electrodes. What is formed at each electrode?
A.Cathode: lead; Anode: bromine
B.Cathode: bromine; Anode: lead
C.Cathode: hydrogen; Anode: oxygen
D.Cathode: lead; Anode: oxygen
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Worked solution
During the electrolysis of molten lead(II) bromide, the positive lead ions migrate to the negative electrode (cathode) where they gain electrons to form lead metal. The negative bromide ions migrate to the positive electrode (anode) where they lose electrons to form bromine gas.
Marking scheme
Award 1 mark for the correct option (A).
Question 32 · multiple_choice
1 marks
Which statement about synthetic polymers is correct?
A.Nylon is formed by addition polymerisation.
B.Poly(ethene) is made from an unsaturated monomer.
C.Polymers are small molecules with low boiling points.
D.Terylene is a naturally occurring polymer.
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Worked solution
Poly(ethene) is formed from ethene, which contains a carbon-carbon double bond and is therefore unsaturated. Nylon and Terylene are synthetic condensation polymers, not addition or naturally occurring.
Marking scheme
Award 1 mark for the correct option (B).
Question 33 · multiple_choice
1 marks
Which row correctly identifies a major source of carbon monoxide and a major source of sulfur dioxide as air pollutants?
A.Carbon monoxide: incomplete combustion of carbon-containing fuels; Sulfur dioxide: combustion of fossil fuels containing sulfur compounds
B.Carbon monoxide: complete combustion of carbon-containing fuels; Sulfur dioxide: car exhaust catalytic converters
C.Carbon monoxide: decomposition of vegetation; Sulfur dioxide: combustion of fossil fuels containing sulfur compounds
D.Carbon monoxide: incomplete combustion of carbon-containing fuels; Sulfur dioxide: respiration in plants
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Worked solution
Carbon monoxide is produced by the incomplete combustion of carbon-containing fuels (such as fossil fuels). Sulfur dioxide is produced by the burning of fossil fuels that contain sulfur compounds as impurities.
Marking scheme
Award 1 mark for the correct option (A).
Question 34 · Multiple Choice
1 marks
A student tests an aqueous solution of an unknown salt X. The addition of aqueous sodium hydroxide to a sample of the solution produces a green precipitate that is insoluble in excess. The addition of dilute nitric acid followed by aqueous silver nitrate to another sample produces a cream precipitate. What is the identity of salt X?
A.iron(III) bromide
B.iron(II) bromide
C.iron(II) chloride
D.copper(II) bromide
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Worked solution
Aqueous sodium hydroxide reacts with iron(II) ions, \( \text{Fe}^{2+} \), to form a green precipitate of iron(II) hydroxide which is insoluble in excess sodium hydroxide. Dilute nitric acid followed by aqueous silver nitrate reacts with bromide ions, \( \text{Br}^- \), to form a cream precipitate of silver bromide. Therefore, the salt is iron(II) bromide.
Marking scheme
1 mark for the correct option B. Reject option A because iron(III) gives a red-brown precipitate. Reject option C because chloride gives a white precipitate. Reject option D because copper(II) gives a light blue precipitate.
Question 35 · Multiple Choice
1 marks
An insoluble salt, lead(II) sulfate, needs to be prepared. Which pair of reactants is most suitable to prepare a pure sample of lead(II) sulfate?
A.lead(II) oxide and dilute sulfuric acid
B.lead(II) nitrate solution and sodium sulfate solution
C.lead(II) carbonate and dilute sulfuric acid
D.lead(II) metal and dilute sulfuric acid
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Worked solution
To prepare an insoluble salt like lead(II) sulfate, precipitation is the most suitable method. This involves mixing two soluble salt solutions. Lead(II) nitrate and sodium sulfate are both soluble in water. When mixed, they react to form a precipitate of insoluble lead(II) sulfate. Reacting lead(II) metal, oxide, or carbonate with sulfuric acid is unsuccessful because an insoluble layer of lead(II) sulfate quickly coats the solid, stopping the reaction.
Marking scheme
1 mark for the correct option B. Reject options A, C, and D because they involve insoluble lead starting materials that will become coated in lead(II) sulfate, preventing the reaction from completing.
Question 36 · Multiple Choice
1 marks
Chlorine gas is bubbled into an aqueous solution of potassium bromide. Which statement correctly describes the observation and explains the reaction?
A.The solution turns brown because chlorine is more reactive than bromine and displaces bromide ions.
B.The solution remains colorless because chlorine is less reactive than bromine and no reaction occurs.
C.The solution turns brown because bromine is more reactive than chlorine and displaces chloride ions.
D.A purple gas is released because chlorine is less reactive than bromine.
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Worked solution
Chlorine is more reactive than bromine because reactivity decreases down Group VII. Therefore, chlorine displaces bromide ions from potassium bromide to form aqueous bromine, which causes the solution to turn brown. The equation for the reaction is \( \text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2 \).
Marking scheme
1 mark for the correct option A. Reject option B because a displacement reaction does occur. Reject option C because chlorine is more reactive than bromine, not less reactive. Reject option D because a purple gas (iodine vapor) is not produced in this reaction.
Question 37 · multiple-choice
1 marks
An experiment is set up to electrolyse molten lead(II) bromide using inert carbon electrodes. Which observations are made at the electrodes?
A.Cathode: bubbles of a colourless gas; Anode: a grey liquid
B.Cathode: a grey liquid; Anode: brown fumes
C.Cathode: brown fumes; Anode: a grey liquid
D.Cathode: a grey liquid; Anode: bubbles of a colourless gas Gold-colored liquid is not formed at either electrode; bromine is a brown gas/vapour at this temperature and lead is a grey liquid metal (molten).
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Worked solution
During the electrolysis of molten lead(II) bromide, \(Pb^{2+}\) ions migrate to the negative electrode (cathode) where they gain electrons to form lead metal, which is observed as a grey liquid. \(Br^{-}\) ions migrate to the positive electrode (anode) where they lose electrons to form bromine gas, which is observed as brown fumes.
Marking scheme
1 mark for correct option B. Reject other options because they swap the products or state incorrect observations.
Question 38 · multiple-choice
1 marks
A student prepares a pure sample of hydrated copper(II) sulfate crystals by reacting excess copper(II) oxide with warm dilute sulfuric acid. Which sequence of steps should the student use after the reaction is complete?
A.evaporate to dryness, filter off the solid, and wash with water
B.filter, evaporate the filtrate to dryness, and wash with acid
C.filter, heat the filtrate to crystallisation point, cool, then filter and dry the crystals
D.evaporate the mixture, cool, filter, and wash the solid with sulfuric acid
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Worked solution
To obtain pure, dry crystals of a soluble salt from a mixture containing an insoluble reactant: 1. Filter the mixture to remove the excess unreacted copper(II) oxide. 2. Heat the filtrate (copper(II) sulfate solution) to evaporate some water until the crystallisation point is reached. 3. Allow the hot saturated solution to cool so crystals can form. 4. Filter the crystals from the remaining solution and dry them (e.g., using filter paper).
Marking scheme
1 mark for option C. Heating to dryness (options A and B) would produce anhydrous powder instead of hydrated crystals. Washing with acid (options B and D) would react with or contaminate the crystals.
Question 39 · multiple-choice
1 marks
Which statement describes the addition polymerisation of ethene to form poly(ethene)?
A.It is an addition reaction where carbon-carbon double bonds open to form single bonds.
B.It is a condensation reaction where water is eliminated.
C.It is an addition reaction where carbon-carbon single bonds open to form double bonds.
D.It is a condensation reaction where carbon-carbon double bonds are formed.
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Worked solution
During the addition polymerisation of ethene, the carbon-carbon double bonds (C=C) in the monomer molecules open up to form carbon-carbon single bonds (C-C) that link the monomer units together in a long chain. No other molecules are formed, so it is not a condensation polymerisation reaction.
Marking scheme
1 mark for option A. Option B and D are incorrect as poly(ethene) formation is an addition polymerisation, not condensation. Option C is incorrect because single bonds do not open to form double bonds.
Question 40 · multiple-choice
1 marks
An aqueous solution of salt \(X\) is tested.
- The addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. - The addition of dilute nitric acid followed by aqueous silver nitrate produces a cream precipitate.
What is the identity of salt \(X\)?
A.iron(II) bromide
B.iron(III) bromide
C.iron(II) chloride
D.copper(II) bromide
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Worked solution
1. **Identify the cation:** The addition of aqueous sodium hydroxide produces a green precipitate that is insoluble in excess. This is the characteristic test result for iron(II) ions, \(\text{Fe}^{2+}\). 2. **Identify the anion:** The addition of dilute nitric acid followed by aqueous silver nitrate produces a cream precipitate. This is the characteristic test result for bromide ions, \(\text{Br}^-\).
Combining these two results, salt \(X\) is iron(II) bromide.
Marking scheme
Award 1 mark for the correct option (A). - Option B is incorrect because iron(III) ions form a red-brown precipitate with aqueous sodium hydroxide. - Option C is incorrect because chloride ions form a white precipitate with silver nitrate. - Option D is incorrect because copper(II) ions form a light blue precipitate with aqueous sodium hydroxide.
Question 41 · multiple-choice
1 marks
Which statement about the relative atomic mass of an element is correct?
A.It is the mass of one atom of the element relative to the mass of a hydrogen atom.
B.It is the average mass of naturally occurring isotopes of the element on a scale where a carbon-12 atom has a mass of exactly 12 units.
C.It is always a whole number representing the sum of protons and neutrons in the most common isotope.
D.It is the mass of one mole of the gaseous element at room temperature and pressure.
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Worked solution
The relative atomic mass of an element is the average mass of its naturally occurring isotopes relative to 1/12th of the mass of a carbon-12 atom.
Marking scheme
Award 1 mark for the correct option (B).
Paper 21 Multiple Choice (Extended)
There are forty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct.
102 Question · 102 marks
Question 1 · Multiple Choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly identifies the product formed at each electrode and the change in pH of the electrolyte surrounding the cathode?
A.Cathode product: hydrogen; Anode product: chlorine; pH at cathode: increases
B.Cathode product: hydrogen; Anode product: chlorine; pH at cathode: decreases
C.Cathode product: sodium; Anode product: chlorine; pH at cathode: increases
D.Cathode product: hydrogen; Anode product: oxygen; pH at cathode: increases
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride:
1. At the cathode, both \(\text{Na}^+\) and \(\text{H}^+\) ions are attracted. Since hydrogen is lower in the reactivity series than sodium, hydrogen ions are preferentially discharged to form hydrogen gas (\(\text{H}_2\)).
2. At the anode, both \(\text{Cl}^-\) and \(\text{OH}^-\) ions are attracted. Because the solution is concentrated, chloride ions are preferentially discharged to form chlorine gas (\(\text{Cl}_2\)).
3. Around the cathode, as hydrogen ions are discharged and leave the solution, the concentration of hydroxide ions (\(\text{OH}^-\)) increases, making the solution alkaline. Thus, the pH of the solution surrounding the cathode increases.
Marking scheme
1 mark for identifying the correct combination: - Cathode product: hydrogen - Anode product: chlorine - pH at cathode: increases
Question 2 · Multiple Choice
1 marks
The structure of a monomer is shown.
\(\text{CH}_2=\text{CH}-\text{COOCH}_3\)
Which formula represents the repeating unit of the addition polymer formed from this monomer?
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Worked solution
In addition polymerisation, the carbon-carbon double bond (\(\text{C}=\text{C}\)) in the monomer breaks to form a single bond (\(\text{C}-\text{C}\)), allowing the monomer units to link together in a continuous chain. The side groups attached to these carbon atoms remain unchanged.
For the monomer \(\text{CH}_2=\text{CH}-\text{COOCH}_3\): - The double bond becomes a single bond: \(-\text{CH}_2-\text{CH}-\) - The side group \(-\text{COOCH}_3\) remains attached to the second carbon atom. - This gives the repeating unit: \(-[\text{CH}_2-\text{CH}(\text{COOCH}_3)]-\).
Marking scheme
1 mark for selecting the correct addition polymer repeating unit showing the saturation of the C=C bond and the preservation of the ester side group (Option A).
Question 3 · Multiple Choice
1 marks
A sample of 1.20 g of magnesium ribbon is reacted completely with an excess of dilute hydrochloric acid.
What is the volume of hydrogen gas produced, measured at room temperature and pressure (r.t.p.)?
[Relative atomic mass: \(\text{Mg} = 24.0\); molar volume of any gas at r.t.p. is \(24.0\text{ dm}^3/\text{mol}\)]
A.\(1.20\text{ dm}^3\)
B.\(2.40\text{ dm}^3\)
C.\(0.12\text{ dm}^3\)
D.\(12.0\text{ dm}^3\)
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Worked solution
1. Calculate the number of moles of magnesium used: \(\text{Moles of Mg} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.20\text{ g}}{24.0\text{ g/mol}} = 0.050\text{ mol}\)
2. Use the stoichiometric ratio from the balanced chemical equation: Since \(1\text{ mol}\) of \(\text{Mg}\) produces \(1\text{ mol}\) of \(\text{H}_2\), then \(0.050\text{ mol}\) of \(\text{Mg}\) will produce \(0.050\text{ mol}\) of \(\text{H}_2\) gas.
3. Calculate the volume of hydrogen gas at r.t.p.: \(\text{Volume} = \text{moles} \times 24.0\text{ dm}^3/\text{mol} = 0.050\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 1.20\text{ dm}^3\).
Marking scheme
1 mark for the correct calculation of the volume of hydrogen gas: 1.20 dm³ (Option A).
Question 4 · Multiple Choice
1 marks
During the electrolysis of concentrated aqueous sodium chloride using inert graphite electrodes, reactions occur at both the anode and the cathode. Which row correctly identifies the substance produced at each electrode and the change in pH of the electrolyte immediately surrounding the cathode?
A.Anode product: chlorine; Cathode product: hydrogen; pH near cathode: increases
B.Anode product: chlorine; Cathode product: sodium; pH near cathode: decreases
C.Anode product: oxygen; Cathode product: hydrogen; pH near cathode: increases
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride (brine), chloride ions (\(\text{Cl}^-\)) are discharged at the anode (positive electrode) to produce chlorine gas. At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) from water are preferentially discharged to produce hydrogen gas, leaving hydroxide ions (\(\text{OH}^-\)) and sodium ions (\(\text{Na}^+\)) in the solution. The concentration of hydroxide ions near the cathode increases, causing the pH in that region to increase (become alkaline).
Marking scheme
1 mark for the correct option (A). No partial marks.
Question 5 · Multiple Choice
1 marks
A student titrates \(25.0\text{ cm}^3\) of \(0.0500\text{ mol/dm}^3\) aqueous sodium hydroxide, \(\text{NaOH}\), with dilute sulfuric acid, \(\text{H}_2\text{SO}_4\).
The volume of sulfuric acid required to neutralize the sodium hydroxide is \(20.0\text{ cm}^3\).
What is the concentration of the sulfuric acid?
A.\(0.0156\text{ mol/dm}^3\)
B.\(0.0313\text{ mol/dm}^3\)
C.\(0.0625\text{ mol/dm}^3\)
D.\(0.125\text{ mol/dm}^3\)
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Worked solution
First, calculate the number of moles of sodium hydroxide used: \(\text{moles of NaOH} = \text{concentration} \times \text{volume} = 0.0500\text{ mol/dm}^3 \times \frac{25.0}{1000}\text{ dm}^3 = 0.00125\text{ mol}\).
According to the balanced chemical equation, \(2\text{ moles of NaOH}\) react with \(1\text{ mole of H}_2\text{SO}_4\). Therefore, the number of moles of sulfuric acid reacted is: \(\text{moles of H}_2\text{SO}_4 = \frac{0.00125}{2} = 0.000625\text{ mol}\).
Next, calculate the concentration of sulfuric acid: \(\text{concentration of H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.000625\text{ mol}}{0.0200\text{ dm}^3} = 0.03125\text{ mol/dm}^3\).
Rounding to 3 significant figures gives \(0.0313\text{ mol/dm}^3\).
Marking scheme
1 mark for the correct option (B). No partial marks.
Question 6 · Multiple Choice
1 marks
Which statement correctly describes addition polymerization or condensation polymerization?
A.Addition polymerization of ethene produces poly(ethene) along with water as a side-product.
B.Condensation polymerization requires monomers with a carbon-carbon double bond, \(\text{C=C}\).
C.The formation of a polyester involves the reaction between a dicarboxylic acid and a diol, which eliminates a small molecule of water.
D.Polyamides are formed by addition polymerization, whereas polyesters are formed by condensation polymerization.
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Worked solution
Addition polymerization involves monomers with a \(\text{C=C}\) double bond joining together without the loss of any other atoms or molecules (no side-products are formed). Condensation polymerization involves monomers with reactive functional groups at both ends, which react to form a polymer chain while releasing a small molecule (usually water). A polyester is formed by the condensation polymerization of a dicarboxylic acid and a diol, eliminating water. Therefore, option C is correct.
Marking scheme
1 mark for the correct option (C). No partial marks.
Question 7 · multiple-choice
1 marks
Two different electrolysis experiments are set up:
- Experiment 1: Electrolysis of concentrated aqueous sodium chloride using inert carbon electrodes. - Experiment 2: Electrolysis of dilute sulfuric acid using inert carbon electrodes.
Which products are formed at the positive electrodes (anodes) in each experiment?
A.Experiment 1: chlorine; Experiment 2: hydrogen
B.Experiment 1: chlorine; Experiment 2: oxygen
C.Experiment 1: hydrogen; Experiment 2: oxygen
D.Experiment 1: sodium; Experiment 2: hydrogen
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Worked solution
In Experiment 1, the electrolyte contains \(\text{Na}^+\), \(\text{Cl}^-\), \(\text{H}^+\), and \(\text{OH}^-\). At the positive electrode (anode), negative ions are attracted. Since the solution is concentrated, halide ions (\(\text{Cl}^-\)) are discharged preferentially over hydroxide ions (\(\text{OH}^-\)) to form chlorine gas (\(\text{Cl}_2\)).
In Experiment 2, the electrolyte contains \(\text{H}^+\), \(\text{SO}_4^{2-}\), and \(\text{OH}^-\). At the positive electrode (anode), negative ions are attracted. Hydroxide ions (\(\text{OH}^-\)) are discharged preferentially over sulfate ions (\(\text{SO}_4^{2-}\)) to form oxygen gas (\(\text{O}_2\)).
- 1 mark for identifying chlorine at the anode of Experiment 1 and oxygen at the anode of Experiment 2. - Incorrect options represent incorrect discharge rules or cathode products.
Question 8 · multiple-choice
1 marks
Which method is most suitable for preparing a pure, dry sample of the insoluble salt, barium sulfate, \(\text{BaSO}_4\)?
A.Mix aqueous barium chloride with dilute sulfuric acid, filter the mixture to collect the precipitate, wash the residue with distilled water, and dry it.
B.Add excess solid barium carbonate to dilute sulfuric acid, filter off the excess solid, and evaporate the filtrate to dryness.
C.Titrate aqueous barium hydroxide with dilute sulfuric acid using an indicator, repeat without the indicator, and evaporate the solution to dryness.
D.Heat barium metal in sulfur dioxide gas, cool the solid residue, and wash it with dilute hydrochloric acid before drying.
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Worked solution
Barium sulfate is an insoluble salt. The standard method for preparing a pure, dry sample of an insoluble salt is precipitation: 1. Mix two solutions containing the required ions, such as soluble barium chloride, \(\text{BaCl}_2(aq)\), and dilute sulfuric acid, \(\text{H}_2\text{SO}_4(aq)\). 2. Filter the mixture to collect the precipitate (barium sulfate residue). 3. Wash the residue on the filter paper with distilled water to remove spectator ions and unreacted soluble impurities. 4. Dry the solid residue in a warm oven or with filter paper.
Option B is incorrect because barium sulfate is insoluble and would coat any unreacted barium carbonate, stopping the reaction, and it would not pass into the filtrate to be crystallized. Option C is incorrect because titration is used for making soluble salts from soluble reactants. Option D is incorrect because it does not produce a pure sample of the salt.
Marking scheme
- 1 mark for selecting the correct precipitation procedure: mixing soluble reagents, filtering, washing the residue with distilled water, and drying.
Question 9 · multiple-choice
1 marks
A condensation polymer is made from a dicarboxylic acid monomer, \(\text{HOOC-CH}_2\text{-COOH}\), and a diol monomer, \(\text{HO-CH}_2\text{-CH}_2\text{-OH}\).
Which structure represents the repeating unit of the polyester formed?
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Worked solution
In condensation polymerization to form a polyester: - The dicarboxylic acid monomer loses its \(\text{-OH}\) groups from both ends: \(\text{-CO-CH}_2\text{-CO-}\) - The diol monomer loses its \(\text{-H}\) atoms from both ends: \(\text{-O-CH}_2\text{-CH}_2\text{-O-}\)
Joining these residues together yields the repeating unit: \(\text{-[-CO-CH}_2\text{-CO-O-CH}_2\text{-CH}_2\text{-O-]-}\)
This is shown in Option A.
Marking scheme
- 1 mark for correctly identifying the ester link and molecular residues formed by the elimination of water molecules.
Question 10 · Multiple Choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly identifies the products formed at each electrode and the change in pH of the electrolyte in the region surrounding the cathode?
A.Anode product: chlorine; Cathode product: hydrogen; pH near cathode: increases
B.Anode product: chlorine; Cathode product: sodium; pH near cathode: decreases
C.Anode product: oxygen; Cathode product: hydrogen; pH near cathode: increases
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Worked solution
In the electrolysis of concentrated aqueous sodium chloride: - At the anode (+), chloride ions (\(Cl^-\)) are discharged in preference to hydroxide ions (\(OH^-\)) because they are in a high concentration, producing chlorine gas: \(2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-\). - At the cathode (-), hydrogen ions (\(H^+\)) are preferentially discharged over sodium ions (\(Na^+\)) because hydrogen is lower in the reactivity series, producing hydrogen gas: \(2H^+(aq) + 2e^- \rightarrow H_2(g)\). - Because \(H^+\) ions are removed from the solution, the concentration of hydroxide ions (\(OH^-\)) near the cathode increases, making the solution alkaline and causing the pH to increase.
Therefore, option A is correct.
Marking scheme
1 mark for the correct option A. - Award 1 mark for identifying chlorine at the anode, hydrogen at the cathode, and an increase in pH near the cathode. - Reject all other combinations.
Question 11 · Multiple Choice
1 marks
How many total ions are present in \(0.10\text{ mol}\) of iron(III) sulfate, \(\text{Fe}_2(\text{SO}_4)_3\)? (The Avogadro constant, \(L\), is \(6.02 \times 10^{23}\text{ mol}^{-1}\).)
A.\(1.20 \times 10^{23}\)
B.\(1.81 \times 10^{23}\)
C.\(3.01 \times 10^{23}\)
D.\(3.01 \times 10^{24}\)
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Worked solution
1. Identify the constituent ions in one formula unit of iron(III) sulfate, \(\text{Fe}_2(\text{SO}_4)_3\): Each formula unit contains 2 \(\text{Fe}^{3+}\) ions and 3 \(\text{SO}_4^{2-}\) ions, giving a total of 5 ions per formula unit.
2. Calculate the number of moles of ions in \(0.10\text{ mol}\) of the compound: \(\text{Moles of ions} = 0.10\text{ mol} \times 5 = 0.50\text{ mol}\).
3. Calculate the actual number of ions using the Avogadro constant: \(\text{Number of ions} = 0.50\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 3.01 \times 10^{23}\).
Therefore, option C is correct.
Marking scheme
1 mark for the correct option C. - Award 1 mark for the correct calculation: \(0.10 \times 5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}\). - Reject option A (only 2 ions counted), option B (only 3 ions counted), and option D (failed to multiply by 0.10).
Which row correctly identifies the types of monomer used to make this polymer and the classification of the polymer?
A.Monomers: diol and dicarboxylic acid; Classification: polyester
B.Monomers: diamine and dicarboxylic acid; Classification: polyamide
C.Monomers: diol and dicarboxylic acid; Classification: polyamide
D.Monomers: diamine and dicarboxylic acid; Classification: polyester
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Worked solution
1. Identify the linkage in the polymer: The polymer chain contains ester linkages (\(-\text{COO}-\)). Thus, it is classified as a polyester. 2. Identify the monomers: A polyester is formed by a condensation reaction between a diol (containing two \(-OH\) groups) and a dicarboxylic acid (containing two \(-COOH\) groups).
Therefore, option A is correct.
Marking scheme
1 mark for the correct option A. - Award 1 mark for identifying the correct monomer types (diol and dicarboxylic acid) and the correct classification (polyester). - Reject options with diamine monomers or polyamide classifications.
Question 13 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly describes the substance produced at the anode and the change in pH of the electrolyte?
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride: At the anode (positive electrode), chloride ions (Cl-) are discharged in preference to hydroxide ions (OH-) because they are in high concentration, producing chlorine gas. At the cathode (negative electrode), hydrogen ions (H+) are discharged in preference to sodium ions (Na+), producing hydrogen gas. As H+ and Cl- ions are removed from the solution, sodium ions (Na+) and hydroxide ions (OH-) remain, forming sodium hydroxide solution. This causes the solution to become alkaline, so the pH increases.
Marking scheme
1 mark for identifying chlorine gas as the product at the anode and that the pH of the remaining solution increases.
Question 14 · multiple-choice
1 marks
A student carries out tests on a green solid, \(X\). When dilute hydrochloric acid is added to solid \(X\), a gas is evolved that turns acidified aqueous potassium manganate(VII) from purple to colourless. When aqueous sodium hydroxide is added to a solution of \(X\), a green precipitate is formed which remains insoluble in excess sodium hydroxide. What is the identity of solid \(X\)?
A.chromium(III) sulfate
B.chromium(III) sulfite
C.iron(II) sulfate
D.iron(II) sulfite
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Worked solution
First, the reaction of solid \(X\) with dilute acid produces a gas that decolourises acidified potassium manganate(VII). This is the characteristic test for sulfur dioxide gas, which is evolved when a sulfite ion \((\text{SO}_3^{2-})\) reacts with dilute acid. Second, when aqueous sodium hydroxide is added to the solution, the formation of a green precipitate that is insoluble in excess confirms the presence of iron(II) ions \((\text{Fe}^{2+})\). (Note that chromium(III) also forms a green precipitate, but it is soluble in excess sodium hydroxide to form a green solution). Therefore, the solid is iron(II) sulfite.
Marking scheme
1 mark for the correct option D.
Question 15 · multiple-choice
1 marks
An aqueous solution contains \(0.10\text{ mol}\) of aluminium sulfate, \(\text{Al}_2(\text{SO}_4)_3\). What is the total number of ions present in this solution? (The Avogadro constant, \(L = 6.02 \times 10^{23}\text{ /mol}\))
A.\(6.02 \times 10^{22}\)
B.\(1.20 \times 10^{23}\)
C.\(3.01 \times 10^{23}\)
D.\(3.01 \times 10^{24}\)
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Worked solution
One formula unit of \(\text{Al}_2(\text{SO}_4)_3\) fully dissociates in solution to yield two \(\text{Al}^{3+}\) ions and three \(\text{SO}_4^{2-}\) ions, which is a total of \(2 + 3 = 5\) ions. Therefore, the total number of moles of ions in the solution is \(0.10\text{ mol} \times 5 = 0.50\text{ mol}\). Using Avogadro's constant, the total number of ions is \(0.50\text{ mol} \times 6.02 \times 10^{23}\text{ /mol} = 3.01 \times 10^{23}\).
Marking scheme
1 mark for the correct option C.
Question 16 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly describes the products formed at each electrode and the change in pH of the electrolyte?
A.product at cathode: hydrogen gas; product at anode: chlorine gas; pH: increases
B.product at cathode: sodium metal; product at anode: chlorine gas; pH: decreases
C.product at cathode: hydrogen gas; product at anode: oxygen gas; pH: increases
D.product at cathode: sodium metal; product at anode: oxygen gas; pH: stays the same
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride: 1. At the cathode (negative electrode), hydrogen ions \((\text{H}^+)\) from water are discharged preferentially over sodium ions because hydrogen is lower in the reactivity series, forming hydrogen gas. 2. At the anode (positive electrode), chloride ions \((\text{Cl}^-)\) are present in high concentration and are discharged preferentially over hydroxide ions, forming chlorine gas. 3. As hydrogen ions and chloride ions are discharged and removed, sodium ions and hydroxide ions remain in solution, forming alkaline sodium hydroxide, which causes the pH to increase.
Marking scheme
1 mark for the correct option A.
Question 17 · multiple-choice
1 marks
An aqueous solution of copper(II) sulfate is electrolysed using two different setups. Setup 1: Copper electrodes are used. Setup 2: Platinum (inert) electrodes are used. Which row correctly describes the observations at the anode (positive electrode) in each setup?
A.Setup 1: Anode decreases in mass. Setup 2: Bubbles of a colourless gas are formed.
B.Setup 1: Bubbles of a colourless gas are formed. Setup 2: Anode decreases in mass.
C.Setup 1: Anode increases in mass. Setup 2: Bubbles of a brown gas are formed.
D.Setup 1: Bubbles of a colourless gas are formed. Setup 2: No change is observed.
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Worked solution
In Setup 1, the copper anode is active and dissolves during electrolysis: \(\text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{e}^-\). Therefore, the anode decreases in mass. In Setup 2, the platinum anode is inert. Hydroxide ions (\(\text{OH}^-\)) are preferentially discharged to form oxygen gas and water: \(4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^-\). Oxygen is a colourless gas, so bubbles of a colourless gas are observed.
Marking scheme
1 mark for selecting the correct option (A).
Question 18 · multiple-choice
1 marks
How many total ions are present in \(200\text{ cm}^3\) of \(0.40\text{ mol/dm}^3\) aqueous iron(III) sulfate, \(\text{Fe}_2(\text{SO}_4)_3\)? (Avogadro constant, \(L = 6.0 \times 10^{23}\text{ /mol}\))
A.\(4.8 \times 10^{22}\)
B.\(9.6 \times 10^{22}\)
C.\(1.4 \times 10^{23}\)
D.\(2.4 \times 10^{23}\)
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Worked solution
First, calculate the number of moles of \(\text{Fe}_2(\text{SO}_4)_3\): \(n = c \times V = 0.40\text{ mol/dm}^3 \times 0.200\text{ dm}^3 = 0.080\text{ mol}\). Each formula unit of \(\text{Fe}_2(\text{SO}_4)_3\) dissociates to produce 5 ions: \(2\text{Fe}^{3+}\) and \(3\text{SO}_4^{2-}\). Total moles of ions = \(0.080\text{ mol} \times 5 = 0.40\text{ mol}\). Total number of ions = \(0.40\text{ mol} \times 6.0 \times 10^{23}\text{ /mol} = 2.4 \times 10^{23}\).
Marking scheme
1 mark for selecting the correct option (D).
Question 19 · multiple-choice
1 marks
A synthetic polymer has the structure shown: \(\text{[-O-CH}_2\text{-CH}_2\text{-O-CO-C}_6\text{H}_4\text{-CO-]_n}\). Which statement about this polymer is correct?
A.It is a polyamide and water is produced during its synthesis.
B.It is a polyester and water is produced during its synthesis.
C.It is a polyamide and no small molecules are produced during its synthesis.
D.It is a polyester and no small molecules are produced during its synthesis.
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Worked solution
The polymer contains the ester linkage \(\text{-O-CO-}\), which classifies it as a polyester. It is formed via condensation polymerisation, during which a small molecule (water) is eliminated/produced as monomers link together.
Marking scheme
1 mark for selecting the correct option (B).
Question 20 · multiple-choice
1 marks
An aqueous solution of copper(II) sulfate is electrolysed using different electrodes. In Experiment 1, inert carbon (graphite) electrodes are used. In Experiment 2, copper electrodes are used. Which row correctly describes the observations at the anode (+ electrode) in each experiment?
A.Experiment 1: Bubbles of a colourless gas are formed. Experiment 2: The electrode decreases in mass.
B.Experiment 1: A pink-brown solid is deposited. Experiment 2: Bubbles of a colourless gas are formed.
C.Experiment 1: Bubbles of a colourless gas are formed. Experiment 2: A pink-brown solid is deposited.
D.Experiment 1: The electrode decreases in mass. Experiment 2: The electrode decreases in mass.
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Worked solution
In Experiment 1, inert carbon electrodes are used. At the anode (positive electrode), hydroxide ions, \( \text{OH}^-(aq) \), from water are discharged in preference to sulfate ions, forming oxygen gas: \( 4\text{OH}^-(aq) \rightarrow \text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \). This is observed as bubbles of a colourless gas. In Experiment 2, copper electrodes are used. Because the copper anode is active, it oxidises and dissolves into the solution: \( \text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + 2e^- \). This is observed as the electrode decreasing in mass. Therefore, option A is correct.
Marking scheme
1 mark for identifying the correct observations at the anode in both experiments: Experiment 1 = bubbles of a colourless gas; Experiment 2 = electrode decreases in mass.
Question 21 · multiple-choice
1 marks
A student titrates \( 25.0\text{ cm}^3 \) of \( 0.0500\text{ mol/dm}^3 \) aqueous sodium hydroxide, \( \text{NaOH} \), with dilute sulfuric acid, \( \text{H}_2\text{SO}_4 \). The equation for the reaction is: \( 2\text{NaOH}(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\text{H}_2\text{O}(l) \) The average titre volume of sulfuric acid required for complete neutralisation is \( 20.0\text{ cm}^3 \). What is the concentration of the sulfuric acid?
A.0.0313 mol/dm³
B.0.0625 mol/dm³
C.0.125 mol/dm³
D.0.0156 mol/dm³
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Worked solution
1. Calculate the number of moles of \( \text{NaOH} \) used: \( \text{Moles of NaOH} = 0.0500\text{ mol/dm}^3 \times 0.0250\text{ dm}^3 = 0.00125\text{ mol} \). 2. Determine the moles of \( \text{H}_2\text{SO}_4 \) reacting using the stoichiometric ratio from the balanced equation (2:1 ratio of NaOH to H2SO4): \( \text{Moles of H}_2\text{SO}_4 = \frac{0.00125}{2} = 0.000625\text{ mol} \). 3. Calculate the concentration of \( \text{H}_2\text{SO}_4 \): \( \text{Concentration} = \frac{0.000625\text{ mol}}{0.0200\text{ dm}^3} = 0.03125\text{ mol/dm}^3 \). Rounding to three significant figures gives \( 0.0313\text{ mol/dm}^3 \). Therefore, option A is correct.
Marking scheme
1 mark for the correct calculation: calculates moles of NaOH as 0.00125 mol, uses the 2:1 mole ratio to find 0.000625 mol of H2SO4, and divides by 0.0200 dm3 to obtain 0.0313 mol/dm3.
Question 22 · multiple-choice
1 marks
A student wishes to prepare a pure, dry sample of the insoluble salt lead(II) sulfate, \( \text{PbSO}_4 \). Which pair of reactants and experimental method should be used?
A.Reactants: aqueous lead(II) nitrate and aqueous sodium sulfate. Method: Filter the mixture, wash the residue with distilled water, and dry.
B.Reactants: insoluble lead(II) carbonate and dilute sulfuric acid. Method: Heat the mixture to evaporate all water, then crystallise.
C.Reactants: aqueous lead(II) nitrate and aqueous sodium sulfate. Method: Heat the mixture to evaporate all the liquid to dryness.
D.Reactants: aqueous lead(II) nitrate and dilute hydrochloric acid. Method: Filter the mixture, wash the residue with distilled water, and dry.
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Worked solution
Lead(II) sulfate is an insoluble salt, which is best prepared by a precipitation reaction using two soluble starting materials. Aqueous lead(II) nitrate and aqueous sodium sulfate are both soluble. Mixing them produces insoluble lead(II) sulfate as a precipitate: \( \text{Pb(NO}_3)_2(aq) + \text{Na}_2\text{SO}_4(aq) \rightarrow \text{PbSO}_4(s) + 2\text{NaNO}_3(aq) \). To obtain a pure, dry sample, the mixture is filtered, the residue of lead(II) sulfate is washed with distilled water to remove spectator ions, and then dried on filter paper. Therefore, option A is correct.
Marking scheme
1 mark for choosing the correct soluble reactants (aqueous lead(II) nitrate and aqueous sodium sulfate) and the correct separation method (filtration, washing the residue with distilled water, and drying).
Question 23 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly identifies the products formed at each electrode and the change in the pH of the remaining solution?
A.Cathode: hydrogen; Anode: chlorine; pH of remaining solution: increases
B.Cathode: sodium; Anode: chlorine; pH of remaining solution: decreases
C.Cathode: hydrogen; Anode: oxygen; pH of remaining solution: remains constant
D.Cathode: sodium; Anode: oxygen; pH of remaining solution: increases
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Worked solution
At the cathode, hydrogen ions ( \( \text{H}^+ \) ) are discharged in preference to sodium ions ( \( \text{Na}^+ \) ) because hydrogen is lower in the reactivity series. This produces hydrogen gas. At the anode, chloride ions ( \( \text{Cl}^- \) ) are discharged in preference to hydroxide ions ( \( \text{OH}^- \) ) because the electrolyte is concentrated, producing chlorine gas. The remaining ions in solution are sodium ( \( \text{Na}^+ \) ) and hydroxide ( \( \text{OH}^- \) ), forming alkaline sodium hydroxide, which causes the pH of the solution to increase.
Marking scheme
Award 1 mark for the correct option A. - Option B is incorrect because sodium is too reactive to discharge at the cathode in aqueous solution. - Option C is incorrect because oxygen is not produced when concentrated chloride solution is used. - Option D is incorrect for both cathode product and anode product.
Question 24 · multiple-choice
1 marks
Hydrogen sulfide reacts with oxygen according to the equation shown:
\( 40\text{ cm}^3 \) of hydrogen sulfide is mixed with \( 80\text{ cm}^3 \) of oxygen. The mixture is ignited and allowed to react. All gas volumes are measured at room temperature and pressure.
What is the total volume of gas remaining after the reaction is complete?
A.\( 40\text{ cm}^3 \)
B.\( 60\text{ cm}^3 \)
C.\( 80\text{ cm}^3 \)
D.\( 100\text{ cm}^3 \)
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Worked solution
According to Avogadro's Law, gas volume ratios are equivalent to their stoichiometric mole ratios.
From the equation, \( 2 \) volumes of \( \text{H}_2\text{S} \) react with \( 3 \) volumes of \( \text{O}_2 \) to produce \( 2 \) volumes of \( \text{SO}_2 \) gas. (Note that water is liquid at room temperature, so its volume is negligible).
1. Determine the limiting reactant: \( 40\text{ cm}^3 \) of \( \text{H}_2\text{S} \) would require: \( 40 \times \frac{3}{2} = 60\text{ cm}^3 \) of \( \text{O}_2 \). Since we have \( 80\text{ cm}^3 \) of \( \text{O}_2 \), hydrogen sulfide is the limiting reactant and is completely consumed.
2. Calculate the volume of excess oxygen remaining: \( 80\text{ cm}^3 - 60\text{ cm}^3 = 20\text{ cm}^3 \) of \( \text{O}_2 \).
3. Calculate the volume of sulfur dioxide gas produced: Since the mole ratio of \( \text{H}_2\text{S} : \text{SO}_2 \) is \( 1 : 1 \), \( 40\text{ cm}^3 \) of \( \text{SO}_2 \) is produced.
Award 1 mark for the correct option B. - Reject A if only the produced sulfur dioxide is counted. - Reject C if the excess reactant calculation is omitted. - Reject D if water is incorrectly treated as a gaseous product.
Question 25 · multiple-choice
1 marks
A synthetic polymer has the structural segment shown:
A.It is a polyamide formed by addition polymerisation.
B.It is a polyester formed by addition polymerisation.
C.It is a polyamide formed by condensation polymerisation.
D.It is a polyester formed by condensation polymerisation.
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Worked solution
The segment contains \( -\text{O}-\text{CO}- \) linkages, which are characteristic of ester linkages. This means the macromolecule is a polyester. Polyesters are formed from dicarboxylic acids and diols with the elimination of water molecules during the reaction. This reaction mechanism is condensation polymerisation.
Marking scheme
Award 1 mark for the correct option D. - Identification of the ester linkage rules out options A and C (polyamides contain amide linkages, \( -\text{NH}-\text{CO}- \)). - Identification of the condensation mechanism rules out option B.
Question 26 · multiple-choice
1 marks
A student wants to prepare a pure, dry sample of copper(II) sulfate crystals starting from the insoluble base copper(II) carbonate. Which sequence of steps should the student follow?
A.Add excess copper(II) carbonate to warm dilute sulfuric acid, filter, heat the filtrate until a saturated solution is formed, allow to cool and crystallise, and dry the crystals.
B.Add dilute sulfuric acid to excess copper(II) carbonate, filter, evaporate all the water from the filtrate to dryness, and dry the solid.
C.Titrate a known volume of dilute sulfuric acid with copper(II) carbonate using an indicator, crystallise, and dry.
D.Add excess dilute sulfuric acid to copper(II) carbonate, filter, wash the residue with water, and dry the residue.
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Worked solution
To prepare a soluble salt from an insoluble base, the insoluble reactant (copper(II) carbonate) is added in excess to ensure all the acid is completely neutralised. The excess unreacted copper(II) carbonate is removed by filtration. To obtain crystals of hydrated copper(II) sulfate, the filtrate is heated to evaporate some of the water until a saturated solution is formed (the crystallisation point). Heating to dryness would produce anhydrous powder instead of hydrated crystals. The solution is then allowed to cool to crystallise, and the crystals are filtered and dried.
Marking scheme
1 mark for the correct option A. Reject option B because evaporating to dryness produces an anhydrous powder, not crystals. Reject option C because titration is only used with soluble reactants. Reject option D because the residue is the unreacted copper(II) carbonate, not the desired copper(II) sulfate salt.
Question 27 · multiple-choice
1 marks
A condensation polymer is formed by reacting a dicarboxylic acid monomer with a diol monomer. Which linkage is formed in this polymer, and which small molecule is eliminated during the reaction?
A.Linkage: ester; Small molecule eliminated: water
B.Linkage: amide; Small molecule eliminated: water
C.Linkage: ester; Small molecule eliminated: hydrogen chloride
D.Linkage: amide; Small molecule eliminated: ammonia
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Worked solution
The reaction of a dicarboxylic acid containing \(-\text{COOH}\) groups with a diol containing \(-\text{OH}\) groups results in the formation of an ester linkage, \(-\text{COO}-\). Because this is a condensation polymerisation reaction, a small water molecule (\(\text{H}_2\text{O}\)) is eliminated for each linkage formed.
Marking scheme
1 mark for identifying the correct ester linkage and the elimination of water (Option A). Amide linkages are found in polyamides (ruling out B and D). Hydrogen chloride is only eliminated when acid chlorides are used, which is outside the standard reactivities defined here (ruling out C).
Question 28 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which products are formed at the electrodes, and how does the pH of the remaining solution change?
A.Product at cathode: hydrogen; Product at anode: chlorine; pH of remaining solution: increases
B.Product at cathode: sodium; Product at anode: chlorine; pH of remaining solution: remains the same
C.Product at cathode: hydrogen; Product at anode: oxygen; pH of remaining solution: decreases
D.Product at cathode: sodium; Product at anode: oxygen; pH of remaining solution: increases
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride: 1. At the cathode (negative electrode), \(\text{H}^+\) ions are preferentially discharged over \(\text{Na}^+\) ions because hydrogen is lower in the reactivity series, forming hydrogen gas (\(\text{H}_2\)). 2. At the anode (positive electrode), \(\text{Cl}^-\) ions are preferentially discharged over \(\text{OH}^-\) ions because they are in a high concentration, forming chlorine gas (\(\text{Cl}_2\)). 3. The remaining ions in solution are \(\text{Na}^+\) and \(\text{OH}^-\), which form sodium hydroxide. This alkaline solution causes the pH of the remaining solution to increase.
Marking scheme
1 mark for the correct combination of hydrogen at the cathode, chlorine at the anode, and an increase in solution pH (Option A). Reject options B and D because sodium is not discharged in aqueous solution. Reject option C because oxygen is not discharged when a concentrated halide solution is used.
Question 29 · Multiple Choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes.
Which row correctly identifies the product at each electrode and the change in pH of the electrolyte immediately surrounding the cathode?
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride:
1. At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) are discharged in preference to sodium ions (\(\text{Na}^+\)) because hydrogen is less reactive than sodium. This produces hydrogen gas: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2(\text{g})\)
2. At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions (\(\text{OH}^-\)) because it is a concentrated solution of a halide. This produces chlorine gas: \(2\text{Cl}^- \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^-
3. Near the cathode, as \)\text{H}^+\) ions are continuously discharged and removed, the relative concentration of hydroxide ions (\(\text{OH}^-\)) increases. The remaining sodium ions (\(\text{Na}^+\)) and hydroxide ions form alkaline sodium hydroxide (\(\text{NaOH}\)) solution, causing the pH of the electrolyte immediately surrounding the cathode to increase.
Marking scheme
1 mark for the correct option A. - Reject B: Sodium is not discharged at the cathode in aqueous solution. - Reject C: Oxygen is not the primary product at the anode for a concentrated chloride solution, and the pH does not remain unchanged. - Reject D: Oxygen is not produced at the cathode.
Question 30 · Multiple Choice
1 marks
A student carries out a titration to find the concentration of a sample of sulfuric acid, \(\text{H}_2\text{SO}_4\).
\(25.0\text{ cm}^3\) of the sulfuric acid is neutralised exactly by \(20.0\text{ cm}^3\) of \(0.150\text{ mol/dm}^3\) sodium hydroxide, \(\text{NaOH}\).
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Worked solution
1. Calculate the moles of \(\text{NaOH}\) used: \(\text{moles of NaOH} = \text{volume in dm}^3 \times \text{concentration} = \frac{20.0}{1000} \times 0.150 = 0.00300\text{ mol}\).
2. Use the stoichiometric ratio from the balanced chemical equation: The molar ratio of \(\text{H}_2\text{SO}_4\) to \(\text{NaOH}\) is \(1:2\). Therefore, \(\text{moles of H}_2\text{SO}_4 = \frac{0.00300}{2} = 0.00150\text{ mol}\).
3. Calculate the concentration of \(\text{H}_2\text{SO}_4\): \(\text{Concentration} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.00150}{25.0 / 1000} = 0.0600\text{ mol/dm}^3\).
Marking scheme
1 mark for the correct option B. - Correct calculation yields 0.0600 mol/dm³. - If the student forgot the 1:2 stoichiometry, they would calculate 0.120 mol/dm³ (option C). - If the student multiplied by 2 instead of dividing by 2, they would calculate 0.240 mol/dm³ (option D). - If the student divided by 4, they would calculate 0.0300 mol/dm³ (option A).
Question 31 · Multiple Choice
1 marks
The diagram shows the structure of a section of a polymer chain.
A.\(\text{HO---CH}_2\text{---CH}_2\text{---OH}\) and \(\text{HOOC---CH}_2\text{---CH}_2\text{---COOH}\)
B.\(\text{HO---CH}_2\text{---CH}_2\text{---OH}\) and \(\text{CH}_3\text{---CH}_2\text{---COOH}\)
C.\(\text{HO---CH}_2\text{---CH}_2\text{---COOH}\) only
D.\(\text{CH}_2=\text{CH}_2\) and \(\text{HOOC---CH}_2\text{---CH}_2\text{---COOH}\)
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Worked solution
The polymer chain contains ester linkages (\(\text{---O---C}(=\text{O})---\)), meaning it is a polyester formed via condensation polymerization.
By breaking the ester linkage: - The alcohol part receives an \(\text{---H}\) atom at each oxygen end to form a diol: \(\text{HO---CH}_2\text{---CH}_2\text{---OH}\). - The carboxylic acid part receives an \(\text{---OH}\) group at each carbonyl carbon end to form a dicarboxylic acid: \(\text{HOOC---CH}_2\text{---CH}_2\text{---COOH}\).
Therefore, these two monomers condense together, losing water (\(\text{H}_2\text{O}\)) molecules, to form the given polymer chain.
Marking scheme
1 mark for the correct option A. - Option B is incorrect because propanoic acid is a mono-carboxylic acid and cannot form a continuous polymer chain. - Option C is incorrect because polymerization of \(\text{HO---CH}_2\text{---CH}_2\text{---COOH}\) would produce a repeating unit with only one carbon chain length between ester links: \(\text{---O---CH}_2\text{---CH}_2\text{---C}(=\text{O})---\). - Option D is incorrect because ethene undergoes addition polymerization, not condensation polymerization with dicarboxylic acids.
Question 32 · multiple-choice
1 marks
A student prepares a pure sample of hydrated copper(II) sulfate crystals, \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}\), from copper(II) oxide and dilute sulfuric acid. Which sequence of steps produces a pure, dry sample of the crystals?
A.Add excess copper(II) oxide to dilute sulfuric acid, filter the mixture, and heat the filtrate to dryness.
B.Add excess copper(II) oxide to dilute sulfuric acid, filter the mixture, heat the filtrate until a hot saturated solution is formed, crystallise, filter, and dry the crystals.
C.Add excess dilute sulfuric acid to copper(II) oxide, filter the mixture, heat the filtrate until a hot saturated solution is formed, crystallise, filter, and dry the crystals.
D.Add excess copper(II) oxide to dilute sulfuric acid, crystallise the mixture directly, filter, and dry the crystals.
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Worked solution
To prepare a pure sample of hydrated copper(II) sulfate crystals: 1. Add excess copper(II) oxide to dilute sulfuric acid to ensure all the acid is neutralised. 2. Filter the mixture to remove the unreacted, excess copper(II) oxide. 3. Heat the filtrate (copper(II) sulfate solution) until it is saturated (the crystallisation point), rather than heating to dryness, to preserve the water of crystallisation. 4. Allow the solution to cool and crystallise, then filter to collect the crystals and dry them with filter paper.
Marking scheme
1 mark for the correct option (B). Reject A: Heating to dryness produces anhydrous copper(II) sulfate powder, not hydrated crystals. Reject C: Using excess acid leaves unreacted acid in the filtrate, contaminating the crystals. Reject D: Crystallising directly without filtering first leaves unreacted copper(II) oxide mixed with the crystals.
Question 33 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes. Which row correctly identifies the product at each electrode and the change in pH of the electrolyte during the electrolysis?
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride: At the cathode (-), hydrogen ions are preferentially discharged over sodium ions because hydrogen is less reactive than sodium, producing hydrogen gas. At the anode (+), chloride ions are preferentially discharged over hydroxide ions because the solution is concentrated, producing chlorine gas. As hydrogen ions and chloride ions are removed, sodium and hydroxide ions remain in the solution, forming sodium hydroxide. This makes the solution alkaline and increases the pH.
Marking scheme
1 mark for the correct option (A). Reject B: Sodium is only discharged in molten electrolysis. Reject C: Chlorine is discharged rather than oxygen due to high chloride concentration. Reject D: Sodium and oxygen are incorrect products for this aqueous mixture.
Question 34 · multiple-choice
1 marks
A sample of \(4.0\text{ g}\) of a gaseous hydrocarbon occupies a volume of \(2.4\text{ dm}^3\) at room temperature and pressure (r.t.p.). [Molar volume of gas at r.t.p. is \(24\text{ dm}^3/\text{mol}\)] What is the molecular formula of the hydrocarbon?
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_3\text{H}_4\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
1. Calculate the number of moles of the hydrocarbon gas: \(\text{moles} = \frac{2.4\text{ dm}^3}{24\text{ dm}^3/\text{mol}} = 0.1\text{ mol}\). 2. Calculate the relative molecular mass (\(M_r\)) of the hydrocarbon: \(M_r = \frac{4.0\text{ g}}{0.1\text{ mol}} = 40\text{ g/mol}\). 3. Determine the molecular formula by calculating the relative molecular mass of each option: \(\text{CH}_4\) is \(16\), \(\text{C}_2\text{H}_4\) is \(28\), \(\text{C}_3\text{H}_4\) is \(40\), and \(\text{C}_3\text{H}_8\) is \(44\). Therefore, the correct formula is \(\text{C}_3\text{H}_4\).
Marking scheme
1 mark for the correct option (C). Reject A, B, and D because their calculated molar masses do not match the experimental molar mass of \(40\text{ g/mol}\).
Question 35 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 36 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 37 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 38 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 39 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 40 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 41 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 42 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 43 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 44 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 45 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 46 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 47 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 48 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 49 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 50 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 51 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 52 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 53 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 54 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 55 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 56 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 57 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 58 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 59 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 60 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 61 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 62 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 63 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 64 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 65 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 66 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 67 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 68 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 69 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 70 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 71 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 72 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 73 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 74 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 75 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 76 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 77 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 78 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 79 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 80 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 81 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 82 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 83 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 84 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 85 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 86 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 87 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 88 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 89 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 90 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 91 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 92 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 93 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 94 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 95 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 96 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 97 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 98 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 99 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 100 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 101 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Question 102 · Multiple Choice
1 marks
A \(0.10\text{ mol}\) sample of a gaseous hydrocarbon, \(X\), is completely burned in an excess of oxygen. This combustion produces \(8.8\text{ g}\) of carbon dioxide, \(\text{CO}_2\), and \(5.4\text{ g}\) of water, \(\text{H}_2\text{O}\). What is the molecular formula of hydrocarbon \(X\)? [Relative atomic masses, \(A_r\): \(\text{H} = 1\); \(\text{C} = 12\); \(\text{O} = 16\)]
A.\(\text{CH}_4\)
B.\(\text{C}_2\text{H}_4\)
C.\(\text{C}_2\text{H}_6\)
D.\(\text{C}_3\text{H}_8\)
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Worked solution
First, calculate the relative formula masses: \(M_r(\text{CO}_2) = 12 + (16 \times 2) = 44\) and \(M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18\). Next, find the moles of products: \(\text{moles of CO}_2 = 8.8 / 44 = 0.20\text{ mol}\) and \(\text{moles of H}_2\text{O} = 5.4 / 18 = 0.30\text{ mol}\). Each mole of \(\text{CO}_2\) contains \(1\text{ mol}\) of \(\text{C}\) atoms, so there are \(0.20\text{ mol}\) of \(\text{C}\). Each mole of \(\text{H}_2\text{O}\) contains \(2\text{ mol}\) of \(\text{H}\) atoms, so there are \(0.30 \times 2 = 0.60\text{ mol}\) of \(\text{H}\). Since the sample contained \(0.10\text{ mol}\) of \(X\), we divide the moles of each element by \(0.10\) to find the formula of 1 mole of the hydrocarbon: \(\text{C} = 0.20 / 0.10 = 2\) and \(\text{H} = 0.60 / 0.10 = 6\). Thus, the molecular formula is \(\text{C}_2\text{H}_6\).
Marking scheme
C is correct. Award 1 mark for calculating the moles of carbon (0.20 mol) and hydrogen (0.60 mol) from the products to determine the C:H ratio in 0.10 mol of the compound as 2:6.
Paper 31 Theory (Core)
Answer all questions. Write your answers in the spaces provided on the question paper. You may use a calculator.
8 Question · 80 marks
Question 1 · Structured
10 marks
A student investigates a mixture of colored food dyes using paper chromatography.
(a) (i) Explain why the start line on the chromatography paper is drawn in pencil rather than ink. [1] (ii) Explain why the level of the solvent in the beaker must be below the start line. [1] (iii) Amino acids can be separated by chromatography but are colorless. State the term for the substance that must be sprayed on the chromatogram to make the amino acids visible. [1] (iv) One of the dyes travels a distance of \(5.2\text{ cm}\) from the start line. The solvent front travels a distance of \(8.0\text{ cm}\) from the start line. Calculate the \(R_f\) value of this dye. Show your working. [2]
(b) Match the correct method of separation from the list below to each of the following mixtures: [3] - Method A: Filtration - Method B: Simple distillation - Method C: Fractional distillation - Method D: Crystallisation
(i) Obtaining pure water from seawater. (ii) Separating sand from a mixture of sand and water. (iii) Separating ethanol from a mixture of ethanol and water.
(c) State how a student could test if a sample of liquid is pure water using its physical properties. [2]
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Worked solution
(a) (i) Pencil is made of graphite, which is insoluble in the solvent and will not dissolve or run. Ink is soluble and would run up the paper with the solvent, interfering with the results. (ii) If the solvent level is above the start line, the spots of dye would dissolve directly into the solvent in the beaker rather than traveling up the paper. (iii) A locating agent is used to colorize colorless spots like amino acids. (iv) The formula is: \(R_f = \frac{\text{distance moved by dye}}{\text{distance moved by solvent}}\) \(R_f = \frac{5.2}{8.0} = 0.65\).
(b) (i) Pure water from seawater requires Simple distillation (Method B) because we want to collect the evaporated solvent. (ii) Sand is an insoluble solid in water, so it is separated by Filtration (Method A). (iii) Ethanol and water are two miscible liquids with different boiling points, so they are separated by Fractional distillation (Method C).
(c) Boil the liquid; pure water has a fixed and sharp boiling point at exactly \(100^\circ\text{C}\).
Marking scheme
(a) (i) Pencil is insoluble / does not run / does not dissolve in the solvent [1] (ii) Prevent the dye from dissolving/washing into the solvent in the beaker [1] (iii) Locating agent [1] (iv) Correct working shown: \(5.2 / 8.0\) [1]; Correct answer: \(0.65\) (accept 0.7, do not accept with units) [1]
(b) (i) Simple distillation / Method B [1] (ii) Filtration / Method A [1] (iii) Fractional distillation / Method C [1]
(c) Measure the boiling point / boil the liquid [1]; pure water boils at exactly \(100^\circ\text{C}\) / constant temperature [1] (or alternative: measure freezing point [1]; pure water freezes at exactly \(0^\circ\text{C}\) [1]).
Question 2 · Structured
10 marks
This question is about atoms, isotopes, and the Periodic Table.
(a) Complete the table below to show the structure of the given particles. [3]
| Particle | Number of protons | Number of neutrons | Number of electrons | | :--- | :---: | :---: | :---: | | Carbon-14 atom (\(^{14}_{6}\text{C}\)) | 6 | **(i)** | 6 | | Sodium ion (\(^{23}_{11}\text{Na}^+\)) | 11 | 12 | **(ii)** | | Oxygen atom (\(^{16}_{8}\text{O}\)) | **(iii)** | 8 | 8 |
(b) Define the term 'isotopes'. [2]
(c) A phosphorus-31 atom is represented by the symbol \(^{31}_{15}\text{P}\). (i) Deduce the number of protons and neutrons in one atom of \(^{31}_{15}\text{P}\). [2] Protons: ______ Neutrons: ______ (ii) State the electronic configuration of a phosphorus atom. [1]
(d) Explain, in terms of electronic configuration, why the elements in Group VIII (Group 0) of the Periodic Table are very unreactive. [2]
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Worked solution
(a) (i) Carbon-14 has a nucleon number of 14 and atomic number of 6. Number of neutrons = \(14 - 6 = 8\). (ii) A neutral sodium atom has 11 electrons. A sodium ion with a \(1+\) charge (\(\text{Na}^+\)) has lost 1 electron: \(11 - 1 = 10\) electrons. (iii) Oxygen's atomic number is 8, which represents the number of protons: 8 protons.
(b) Isotopes are atoms of the same element with the same number of protons (or atomic number) but different numbers of neutrons (or nucleon / mass numbers).
(c) (i) For \(^{31}_{15}\text{P}\), the atomic number (bottom) is 15, which is the number of protons. The nucleon number (top) is 31, so the number of neutrons is \(31 - 15 = 16\). (ii) A phosphorus atom has 15 electrons. The electronic configuration is filled as 2 in the first shell, 8 in the second, and 5 in the third: 2,8,5.
(d) Elements in Group VIII have a stable, full outer shell of electrons. Because their outer shell is complete, they do not need to lose, gain, or share electrons with other atoms to react.
Marking scheme
(a) (i) 8 [1] (ii) 10 [1] (iii) 8 [1]
(b) Atoms with the same proton number / atomic number [1]; but different nucleon number / mass number / number of neutrons [1]
(d) They have a full outer shell of electrons / complete outer shell [1]; so they are stable / do not need to gain, lose, or share electrons [1]
Question 3 · Structured
10 marks
Electrolysis is a useful process for breaking down ionic substances using electricity.
(a) (i) State what is meant by the term 'electrolysis'. [2] (ii) Name a suitable material that can be used for inert electrodes in a school laboratory. [1]
(b) Lead(II) bromide, \(\text{PbBr}_2\), is electrolysed using inert electrodes. (i) Explain why solid lead(II) bromide does not conduct electricity, whereas molten lead(II) bromide does. [2] (ii) Describe the observation and name the product formed at the negative electrode (cathode) during the electrolysis of molten lead(II) bromide. [2] Observation: ______ Product name: ______ (iii) Describe the observation and name the product formed at the positive electrode (anode) during the electrolysis of molten lead(II) bromide. [2] Observation: ______ Product name: ______
(c) During the industrial extraction of aluminium by electrolysis, state the name of the main ore from which aluminium oxide is obtained. [1]
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Worked solution
(a) (i) Electrolysis is the breakdown of an ionic compound, either molten or in aqueous solution, by the passage of an electric current. (ii) Inert electrodes are made of materials that do not react during electrolysis. Common examples are graphite (carbon) or platinum.
(b) (i) In solid lead(II) bromide, the ions are held tightly in a fixed ionic lattice and cannot move to carry the current. When molten, the ionic lattice is broken, and the ions are free to move and conduct electricity. (ii) At the negative electrode (cathode), lead ions (\(\text{Pb}^{2+}\)) gain electrons to form lead metal. The observation is a grey liquid or grey metallic beads. The product name is lead. (iii) At the positive electrode (anode), bromide ions (\(\text{Br}^{-}\)) lose electrons to form bromine gas. The observation is a brown gas or brown fumes. The product name is bromine.
(c) The main ore of aluminium is bauxite.
Marking scheme
(a) (i) Breakdown of an ionic compound / substance [1]; by the passage of electricity / electric current [1] (ii) Graphite / carbon / platinum [1]
(b) (i) In solid, the ions are in fixed positions / cannot move [1]; in molten, the ions are free to move [1] (Note: Mention of 'electrons' moving in the molten salt scores 0 for that part) (ii) Observation: grey liquid / grey beads / grey solid [1]; Product: lead [1] (iii) Observation: brown gas / brown fumes [1]; Product: bromine (do not accept bromide) [1]
(c) Bauxite [1]
Question 4 · Structured
10 marks
This question is about the Group VII elements and their structures.
(a) State the group number of the halogens. [1]
(b) Describe the trend in reactivity of the Group VII elements down the group. [1]
(c) Fluorine and chlorine are in Group VII. (i) State the number of protons, neutrons, and electrons in an atom of fluorine, \(^{19}_{9}\text{F}\). [3] (ii) A chloride ion has the formula \(\text{Cl}^-\). State the number of electrons in a \(\text{Cl}^-\) ion. (The proton number of chlorine is 17). [1]
(d) Chlorine reacts with aqueous potassium bromide. (i) State the observations for this reaction. [2] (ii) Complete the word equation for this reaction: chlorine + potassium bromide \(\rightarrow\) .................... + .................... [2]
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Worked solution
(a) Halogens belong to Group VII (or Group 7 / 17).
(b) As you go down Group VII, the chemical reactivity of the elements decreases.
(c)(i) For \(^{19}_{9}\text{F}\): - Number of protons = atomic (proton) number = 9. - Number of neutrons = mass number - atomic number = 19 - 9 = 10. - Number of electrons in a neutral atom = number of protons = 9.
(c)(ii) A neutral chlorine atom has 17 electrons. A chloride ion (\(\text{Cl}^-\)) has gained one electron, so it has 17 + 1 = 18 electrons.
(d)(i) Chlorine is more reactive than bromine, so it displaces bromine from potassium bromide. The resulting bromine gas/solution makes the solution turn orange-brown from colourless.
(d)(ii) Chlorine reacts with potassium bromide to produce potassium chloride and bromine.
(d)(i) 2 marks: - turns orange / yellow / brown / red-brown [1] - from colourless [1]
(d)(ii) 2 marks: - potassium chloride [1] - bromine [1] (Accept products in either order)
Question 5 · Structured
10 marks
This question is about electrolysis.
(a) Complete the sentences about the electrolysis of molten lead(II) bromide using words from the list. Each word may be used once, more than once, or not at all.
During electrolysis, positive ions move to the negative electrode, which is called the .................... . At this electrode, the metal .................... is formed. Negative ions move to the positive electrode, which is called the .................... . At this electrode, the non-metal .................... is formed. [4]
(b) State the observations at each electrode during the electrolysis of molten lead(II) bromide: (i) positive electrode (anode) [1] (ii) negative electrode (cathode) [1]
(c) Inert electrodes are used in this process. State the name of a material suitable for making inert electrodes. [1]
(d) When aqueous copper(II) sulfate is electrolysed using carbon electrodes: (i) State the product formed at the cathode. [1] (ii) State the product formed at the anode. [1] (iii) Describe how the appearance of the solution changes during this electrolysis. [1]
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Worked solution
(a) - Positive ions (cations) move to the negative electrode, which is called the cathode. - At the cathode, metal ions (\(\text{Pb}^{2+}\)) are reduced to lead metal. - Negative ions (anions) move to the positive electrode, which is called the anode. - At the anode, non-metal ions (\(\text{Br}^{-}\)) are oxidised to bromine.
(b)(i) Bromine gas is produced at the anode, which is observed as a brown gas/vapour. (b)(ii) Molten lead is produced at the cathode, observed as a shiny/silvery grey liquid puddle at the bottom of the electrode.
(c) Carbon (in the form of graphite) or platinum are commonly used as inert electrodes because they conduct electricity well but do not react.
(d)(i) Copper is less reactive than hydrogen, so copper metal is deposited at the cathode. (d)(ii) Hydroxide ions from water are oxidised at the anode, releasing oxygen gas. (d)(iii) As the blue copper(II) ions are removed from the solution to form copper metal at the cathode, the blue colour of the solution gradually fades/becomes lighter.
(a) State which compound, A, B, C or D: (i) is an alkene. [1] (ii) is an alcohol. [1] (iii) contains a carboxylic acid functional group. [1] (iv) is a saturated hydrocarbon. [1]
(b) Describe a chemical test to distinguish between Compound A and Compound B. test: .................... result with Compound A: .................... result with Compound B: .................... [3]
(c) Compound B (ethene) can be polymerised to form poly(ethene). (i) State the type of polymerisation reaction that occurs. [1] (ii) Complete the sentence: Ethene is the .................... in this reaction, and poly(ethene) is the polymer. [1] (iii) State one common use of poly(ethene). [1]
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Worked solution
(a)(i) Compound B (ethene, \(\text{C}_2\text{H}_4\)) contains a carbon-carbon double bond, so it is an alkene. (a)(ii) Compound C (ethanol, \(\text{C}_2\text{H}_5\text{OH}\)) contains the \(-\text{OH}\) functional group, so it is an alcohol. (a)(iii) Compound D (ethanoic acid, \(\text{CH}_3\text{COOH}\)) contains the \(-\text{COOH}\) group, so it is a carboxylic acid. (a)(iv) Compound A (methane, \(\text{CH}_4\)) contains only carbon-carbon single bonds and is a hydrocarbon, so it is a saturated hydrocarbon.
(b) Bromine water (aqueous bromine) is used to test for unsaturation. The unsaturated alkene (Compound B) reactively decolourises the bromine water, whereas the saturated alkane (Compound A) does not react in the absence of UV light, so the solution remains orange-brown.
(c)(i) Alkenes undergo addition polymerisation to form polymers. (c)(ii) The small repeating unit molecules that join together are called monomers. (c)(iii) Poly(ethene) is widely used to make plastic bags, bottles, toys, packaging film, and electrical wire insulation.
Marking scheme
(a) 4 marks: - (i) B [1] - (ii) C [1] - (iii) D [1] - (iv) A [1]
(b) 3 marks: - test: aqueous bromine / bromine water / bromine [1] - result with A: remains orange / brown / yellow / no change [1] - result with B: decolourises / turns colourless [1] (reject: turns clear)
(a) A student prepares a sample of soluble copper(II) sulfate crystals by reacting insoluble copper(II) oxide with dilute sulfuric acid. Write the word equation for this reaction. [1]
(b) The student adds excess copper(II) oxide to the warm dilute sulfuric acid. Explain why excess copper(II) oxide is added. [1]
(c) Describe how the excess copper(II) oxide is removed from the reaction mixture. [1]
(d) Describe the steps needed to obtain pure, dry crystals of hydrated copper(II) sulfate from the filtrate. [3]
(e) In a different experiment, the student prepares a sample of the insoluble salt, barium sulfate. (i) Suggest the names of two soluble salts that can be reacted together to form barium sulfate. [2] (ii) State the name of this type of reaction. [1] (iii) State how the solid barium sulfate is separated from the reaction mixture. [1]
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(b) To ensure all of the sulfuric acid is completely reacted and neutralised.
(c) Filtration / filtering the mixture.
(d) 1. Heat the filtrate until the crystallisation point is reached (to obtain a saturated solution). 2. Leave the solution to cool and crystallise. 3. Filter off the crystals and dry them with filter paper or in a warm oven.
(b) [1 mark] - To ensure all the acid is reacted / used up / neutralised (Reject: to speed up reaction).
(c) [1 mark] - Filtration / filtering.
(d) [3 marks] - M1: Heat / evaporate the filtrate to crystallisation point / until saturated [1] (Reject: evaporate to dryness). - M2: Cool / leave to cool (to allow crystals to form) [1]. - M3: Filter the crystals AND dry them (with filter paper / in a warm oven) [1].
(a) Poly(ethene) is a widely used synthetic addition polymer. (i) What is meant by the term polymer? [1] (ii) Ethene is the monomer used to make poly(ethene). State the type of covalent bond present in ethene that is broken during the polymerisation process. [1] (iii) State one common use of poly(ethene). [1]
(b) Propene is another alkene monomer. It polymerises to form poly(propene). Complete the table below to compare the properties of the monomer propene with the polymer poly(propene).
| Property | Monomer: Propene | Polymer: Poly(propene) | | :--- | :--- | :--- | | State of matter at room temperature | (i) .................... [1] | (ii) .................... [1] | | Classification (Saturated or Unsaturated) | (iii) .................... [1] | (iv) .................... [1] |
(c) Starch is a natural polymer. (i) Name the type of monomer unit that links together to form starch. [1] (ii) Name one other natural polymer found in living organisms. [1] (iii) State the process by which complex carbohydrates like starch are broken down into simple sugars in the presence of an acid catalyst. [1]
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Worked solution
(a)(i) A large molecule built up from many small units (monomers) joined together. (a)(ii) Carbon-carbon double bond / \(C=C\) double bond. (a)(iii) Plastic bags / plastic bottles / toys / packaging / cling film.
(b) (i) gas (ii) solid (iii) unsaturated (iv) saturated
Answer all questions. Write your answers in the spaces provided on the question paper. You should show all your working and use appropriate units.
6 Question · 78.99 marks
Question 1 · Structured / Short Answer
13 marks
A student carries out a titration to determine the concentration of a solution of sulfuric acid, \(\text{H}_2\text{SO}_4\).
They titrate \(25.0\text{ cm}^3\) of \(0.150\text{ mol/dm}^3\) sodium hydroxide, \(\text{NaOH}\), with the sulfuric acid. The equation for the reaction is:
(a) Describe how the student would carry out this titration. In your answer, include: - the names of the apparatus used to measure the volume of \(\text{NaOH}\) and to add the acid - how the student knows the end-point is reached using a named indicator and its color change. [5 marks]
(b) The average volume of sulfuric acid used in the titration is \(18.75\text{ cm}^3\).
(i) Calculate the number of moles of \(\text{NaOH}\) in \(25.0\text{ cm}^3\) of \(0.150\text{ mol/dm}^3\) solution. [1 mark]
(ii) Determine the number of moles of \(\text{H}_2\text{SO}_4\) that reacted with this amount of \(\text{NaOH}\). [1 mark]
(iii) Calculate the concentration of the sulfuric acid in \(\text{mol/dm}^3\). [2 marks]
(iv) Calculate the concentration of the sulfuric acid in \(\text{g/dm}^3\).
(c) After the titration, the student wants to obtain pure, dry crystals of hydrated sodium sulfate, \(\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}\).
Describe the steps they should take, starting with the reaction mixture from a fresh titration performed without the indicator. [2 marks]
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Worked solution
(a) 1. Measure \(25.0\text{ cm}^3\) of sodium hydroxide using a **pipette** (and pipette filler). 2. Transfer the sodium hydroxide to a **conical flask**. 3. Add a few drops of a suitable indicator, e.g. **phenolphthalein** (solution turns **pink**). 4. Fill a **burette** with sulfuric acid, then add the acid to the flask while **swirling** constantly. 5. Stop adding acid when the indicator permanently changes color (for phenolphthalein, from **pink to colorless**).
(ii) From the equation, 1 mole of \(\text{H}_2\text{SO}_4\) reacts with 2 moles of \(\text{NaOH}\). \(\text{Moles of H}_2\text{SO}_4 = \frac{0.00375}{2} = 0.001875\text{ mol}\)
(iii) \(\text{Concentration of H}_2\text{SO}_4 = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.001875}{18.75 / 1000} = 0.100\text{ mol/dm}^3\)
(c) 1. Heat the solution to evaporate water until the crystallization point is reached (or until saturated). 2. Allow the hot saturated solution to cool slowly so that crystals of \(\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}\) form. 3. Filter the crystals from the remaining solution, wash them with a small amount of cold distilled water, and dry them using filter paper.
Marking scheme
(a) [5 marks total] - M1: Name of **pipette** used to measure the volume of NaOH [1] - M2: Name of **burette** used to add the acid [1] - M3: Name of a suitable indicator, e.g. **phenolphthalein** (or methyl orange) [1] - M4: Correct color change associated with the chosen indicator at the end-point (e.g. **pink to colorless** for phenolphthalein; **yellow to orange/red** for methyl orange) [1] - M5: Detailed practical action, e.g. **swirling** the flask during addition, or adding acid **dropwise** near the end-point [1]
(b) [6 marks total] - (i) 0.00375 mol [1] - (ii) 0.001875 mol (or error carried forward from (i) divided by 2) [1] - (iii) 0.100 mol/dm³ (or error carried forward: moles from (ii) / 0.01875) [2] - Give 1 mark for correct formula or substitution: Conc = moles / 0.01875. - Give 1 mark for correct final value. - (iv) 9.80 g/dm³ (or error carried forward: concentration from (iii) * 98) [2] - Give 1 mark for calculating Mr(H2SO4) = 98. - Give 1 mark for correct final value.
(c) [2 marks total] - M1: **Evaporate/heat** the solution to crystallization point / until crystals start to form on a glass rod [1] - M2: **Filter** the crystals, **wash** with cold distilled water, and **dry** (e.g. with filter paper) [1] (All three actions needed for the mark; do not accept heating to dryness).
Question 2 · Structured / Short Answer
13 marks
Electrolysis is a key chemical process used to extract metals, produce important chemicals, and electroplate objects.
(a) Molten lead(II) bromide is electrolysed using inert carbon electrodes.
(i) State the observation made at each electrode. [2 marks] - Cathode (negative electrode): - Anode (positive electrode):
(ii) Write the ionic half-equation, including state symbols, for the reaction occurring at the anode. [2 marks]
(b) Concentrated aqueous sodium chloride (brine) is electrolysed using inert electrodes.
(i) Identify the gaseous product formed at each electrode. [2 marks] - Cathode: - Anode:
(ii) Explain why sodium metal is not produced at the cathode. [1 mark]
(iii) After electrolysis has occurred for some time, the electrolyte is tested with universal indicator and found to be alkaline. Identify the ion responsible for this alkalinity and explain how it is formed during this electrolysis. [2 marks]
(c) Electroplating is used to coat a steel spoon with nickel.
(i) State what should be used as: - the anode (positive electrode) [1 mark] - the cathode (negative electrode) [1 mark] - the electrolyte [1 mark]
(ii) Write the ionic half-equation for the reaction occurring at the cathode during this electroplating. [1 mark]
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Worked solution
(a) (i) - Cathode: **Grey liquid / shiny bead** of metal (lead) is formed at the bottom of the tube. - Anode: **Brown fumes / gas** (bromine) are evolved.
(ii) Sodium is highly reactive, meaning sodium ions are much more stable than hydrogen ions. Consequently, **hydrogen ions are preferentially reduced/discharged** at the cathode instead of sodium ions.
(iii) - Ion responsible: **Hydroxide** / \(\text{OH}^-\) - Explanation: Water dissociates into \(\text{H}^+\) and \(\text{OH}^-\) ions. As \(\text{H}^+\) ions are discharged at the cathode (as hydrogen gas) and \(\text{Cl}^-\) ions are discharged at the anode (as chlorine gas), the \(\text{Na}^+\) and \(\text{OH}^-\) ions remain in the solution, increasing the concentration of \(\text{OH}^-\) ions (forming sodium hydroxide, which is alkaline).
(c) (i) - Anode: **Nickel** (metal sheet/strip) - Cathode: The **steel spoon** - Electrolyte: A soluble nickel salt solution, e.g., **nickel(II) sulfate** solution / \(\text{NiSO}_4\text{(aq)}\) (or nickel chloride)
(ii) \(\text{Ni}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Ni(s)}\)
Marking scheme
(a) [4 marks total] - (i) Cathode: grey liquid / shiny silver liquid (accept grey solid) [1]; Anode: brown gas / brown vapour [1]. - (ii) 2Br-(l) -> Br2(g) + 2e- [2] - 1 mark for correct species and balancing. - 1 mark for correct state symbols (must show liquid reactant and gaseous product).
(b) [5 marks total] - (i) Cathode: hydrogen [1]; Anode: chlorine [1]. - (ii) Hydrogen ions are lower in the reactivity series than sodium ions / hydrogen ions are preferentially discharged / sodium is more reactive than hydrogen [1]. - (iii) Hydroxide ions / OH- [1]; Hydrogen and chloride ions are discharged / removed, leaving sodium and hydroxide ions in solution [1].
(c) [4 marks total] - (i) Anode: nickel [1]; Cathode: steel spoon [1]; Electrolyte: nickel sulfate solution / nickel chloride solution (accept any soluble nickel(II) salt, reject just 'nickel' or 'nickel solution') [1]. - (ii) Ni2+(aq) + 2e- -> Ni(s) [1] (ignore state symbols here, but must be chemically correct).
Question 3 · Structured / Short Answer
13 marks
Polymers are macromolecules built up from small monomer units. They can be classified as addition or condensation polymers.
(a) Propene, \(\text{CH}_3\text{CH}=\text{CH}_2\), is an unsaturated hydrocarbon that can undergo addition polymerisation to form poly(propene).
(i) Draw a diagram to show the structure of a propene monomer. Show all atoms and all bonds. [1 mark]
(ii) Draw the structure of poly(propene) showing two repeat units. Show all atoms and all single bonds. [2 marks]
(iii) State the main difference between addition polymerisation and condensation polymerisation in terms of the products formed. [1 mark]
(b) Nylon is a synthetic polyamide formed by condensation polymerisation.
(i) Nylon can be made from a dicarboxylic acid and a diamine. Draw the structures of a dicarboxylic acid monomer and a diamine monomer. Use block diagrams (rectangles) to represent the carbon chains. Show the functional groups at each end. [2 marks] - Dicarboxylic acid monomer: - Diamine monomer:
(ii) Draw the structure of the amide linkage (peptide link) showing all atoms and bonds. [1 mark]
(iii) Identify the small molecule that is eliminated during this polymerisation. [1 mark]
(c) Terylene is a polyester.
(i) State the name of the two functional groups that react to form a polyester. [2 marks]
(ii) Draw the structure of a polyester linkage (ester link) showing all atoms and bonds. [1 mark]
(d) Many synthetic addition polymers are non-biodegradable.
(i) Explain what is meant by the term non-biodegradable and state one environmental problem associated with the disposal of these polymers in landfills. [2 marks]
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Worked solution
(a) (i) Propene monomer structure: ``` H H H | | | H - C = C - C - H | H ``` (The double bond between C1 and C2 must be shown, and C3 has three single bonds to H).
(ii) Poly(propene) structure (two repeat units): ``` H H H H | | | | - - C - C - C - C - - | | | | H CH3 H CH3 ``` (All bonds in the backbone must be single, with continuation bonds shown at both ends, and the methyl groups clearly displayed as side chains).
(iii) Addition polymerisation produces **only the polymer** as the sole product, whereas condensation polymerisation produces the polymer **and a small molecule** (such as water or hydrogen chloride) as a byproduct.
(ii) Amide linkage: ``` O H || | - C - N - ``` (All atoms C, O, N, H and their bonds must be shown: C=O double bond and N-H single bond).
(iii) **Water** / \(\text{H}_2\text{O}\)
(c) (i) **Carboxylic acid** group (or carboxyl) and **alcohol** group (or hydroxyl / diol).
(ii) Ester linkage: ``` O || - C - O - ``` (C=O double bond and C-O single bond must be shown).
(d) (i) - **Non-biodegradable**: Cannot be decomposed / broken down by microorganisms or decomposers (bacteria/fungi). - **Environmental problem**: It takes up valuable space in landfill sites (fills up landfills), causes visual pollution, or can harm/trap wildlife.
Marking scheme
(a) [4 marks total] - (i) Correct fully displayed structure of propene showing all atoms and bonds (including C=C double bond and C-H single bonds) [1]. - (ii) Correct structure showing two repeat units with a single-bonded carbon backbone and methyl side chains, with continuation bonds at both ends [2]. - 1 mark for correct repeat unit backbone with single bonds. - 1 mark for correct attachment of groups (one H and one -CH3 on alternating carbons) and continuation bonds. - (iii) Addition polymerisation forms **only one product** (the polymer) / condensation polymerisation also produces **a small molecule** (e.g. water) [1].
(b) [4 marks total] - (i) - Correct dicarboxylic acid monomer structure showing both -COOH groups attached to a block [1]. - Correct diamine monomer structure showing both -NH2 groups attached to a block [1]. - (ii) Correctly drawn amide linkage showing -C(=O)-NH- with all bonds displayed [1]. - (iii) **Water** / H2O (or hydrogen chloride / HCl) [1].
(c) [3 marks total] - (i) **Carboxylic acid** [1] and **alcohol** / diol [1] (accept carboxyl and hydroxyl; do not accept hydroxide). - (ii) Correctly drawn ester linkage showing -C(=O)-O- with all bonds displayed [1].
(d) [2 marks total] - (i) - Definition: Cannot be broken down/decomposed by microbes/bacteria/fungi [1]. - Landfill issue: Accumulates in landfills / takes up space / harms wildlife / visual pollution [1] (Do not accept 'releases toxic gases' unless specifically linked to incineration, which is not a landfill issue).
Question 4 · Structured
13.33 marks
A student titrates a \( 25.0\text{ cm}^3 \) sample of a diprotic acid, \( \text{H}_2\text{A} \), of unknown concentration against a standard \( 0.150\text{ mol/dm}^3 \) solution of sodium hydroxide, \( \text{NaOH} \). (a) Describe how the student should rinse and prepare the burette before filling it with the standard \( \text{NaOH} \) solution. (b) State the colour change of methyl orange indicator at the end point when the alkali is added to the acid in the conical flask. (c) The student performs three titrations and obtains the following titres: Titration 1: \( 23.40\text{ cm}^3 \), Titration 2: \( 22.80\text{ cm}^3 \), Titration 3: \( 22.90\text{ cm}^3 \). Calculate the average concordant titre to be used in subsequent calculations. (d) Write the balanced chemical equation for the reaction between \( \text{H}_2\text{A} \) and \( \text{NaOH} \). (e) Calculate the concentration of the diprotic acid, \( \text{H}_2\text{A} \), in \( \text{mol/dm}^3 \). (f) Given that the concentration of \( \text{H}_2\text{A} \) is \( 3.42\text{ g/dm}^3 \), calculate the relative molecular mass, \( M_r \), of the acid. Show your working.
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Worked solution
(a) Rinse the burette first with distilled water and then with the \( 0.150\text{ mol/dm}^3\text{ NaOH} \) solution to prevent dilution. (b) The colour change of methyl orange when alkali is added to acid is from red/pink to yellow (or orange). (c) Concordant titres are within \( 0.10\text{ cm}^3 \) of each other, which are Titration 2 and Titration 3. Average titre = \( (22.80 + 22.90) / 2 = 22.85\text{ cm}^3 \). (d) Balanced equation: \( \text{H}_2\text{A} + 2\text{NaOH} \rightarrow \text{Na}_2\text{A} + 2\text{H}_2\text{O} \). (e) Moles of \( \text{NaOH} = (22.85 / 1000) \times 0.150 = 3.4275 \times 10^{-3}\text{ mol} \). From the equation stoichiometry, moles of \( \text{H}_2\text{A} = (3.4275 \times 10^{-3}) / 2 = 1.7138 \times 10^{-3}\text{ mol} \). Concentration of \( \text{H}_2\text{A} = (1.7138 \times 10^{-3}\text{ mol}) / 0.0250\text{ dm}^3 = 0.06855\text{ mol/dm}^3 \) (rounds to \( 0.0686\text{ mol/dm}^3 \)). (f) \( M_r = \text{concentration in g/dm}^3 / \text{concentration in mol/dm}^3 = 3.42 / 0.06855 = 49.9 \).
Marking scheme
(a) [2 marks]: 1 mark for rinsing with distilled water; 1 mark for rinsing with sodium hydroxide solution. (b) [1 mark]: Red to yellow / orange. (c) [2 marks]: 1 mark for identifying the concordant titres (22.80 and 22.90); 1 mark for calculating the correct average of 22.85. (d) [2 marks]: 1 mark for correct formulas of reactants and products; 1 mark for balancing. (e) [3 marks]: 1 mark for calculating moles of NaOH; 1 mark for using the 1:2 ratio to find moles of H2A; 1 mark for final concentration in mol/dm3 (accept 0.0685 to 0.0687). (f) [3 marks]: 1 mark for the relationship Mr = mass/moles; 1 mark for dividing 3.42 by their answer in (e); 1 mark for correct final Mr value (accept 49.7 to 50.0 based on rounding).
Question 5 · Structured
13.33 marks
Aqueous copper(II) sulfate, \( \text{CuSO}_4\text{(aq)} \), is electrolysed. (a) (i) Describe the observation at the cathode and write an ionic half-equation for the reaction occurring at this electrode when inert carbon electrodes are used. (ii) Describe the observation at the anode and write an ionic half-equation for the reaction occurring at this electrode when inert carbon electrodes are used. (b) Explain why the blue colour of the electrolyte fades and the solution eventually becomes acidic during this electrolysis with carbon electrodes. (c) The experiment is repeated using copper electrodes instead of carbon electrodes. (i) Describe the change in mass of the anode and cathode. (ii) Explain how this electrolysis setup with copper electrodes is adapted and used in the industrial purification of copper.
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Worked solution
(a)(i) Cathode: A pink/brown solid (copper) is deposited. Half-equation: \( \text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu}\text{(s)} \). (a)(ii) Anode: Bubbles of a colourless gas (oxygen) are formed. Half-equation: \( 4\text{OH}^-\text{(aq)} \rightarrow \text{O}_2\text{(g)} + 2\text{H}_2\text{O}\text{(l)} + 4\text{e}^- \). (b) The blue colour fades because \( \text{Cu}^{2+} \) ions are discharged/removed from the solution. The solution becomes acidic because \( \text{OH}^- \) ions are discharged leaving excess \( \text{H}^+ \) ions (from water) and \( \text{SO}_4^{2-} \) ions in solution, forming sulfuric acid. (c)(i) The anode mass decreases (dissolves) and the cathode mass increases. (c)(ii) The anode is made of impure copper and the cathode is made of pure copper. Copper atoms at the impure anode lose electrons to form copper ions, which dissolve into the electrolyte. These copper ions migrate to the cathode, where they gain electrons and deposit as pure copper metal, leaving impurities behind as anode slime.
Marking scheme
(a)(i) [2 marks]: 1 mark for pink/brown solid/deposit; 1 mark for correct balanced half-equation. (a)(ii) [2 marks]: 1 mark for bubbles of colourless gas; 1 mark for correct balanced half-equation. (b) [3 marks]: 1 mark for stating that blue colour fades due to the removal of Cu2+ ions; 1 mark for stating that OH- is discharged; 1 mark for explaining that excess H+ and SO42- remain to form acid. (c)(i) [2 marks]: 1 mark for anode mass decreases; 1 mark for cathode mass increases. (c)(ii) [4 marks]: 1 mark for identifying impure copper anode and pure copper cathode; 1 mark for oxidation of copper at the anode; 1 mark for reduction of copper ions at the cathode; 1 mark for impurities falling to the bottom / leaving pure copper at cathode.
Question 6 · Structured
13.33 marks
Polymers can be synthesized through addition or condensation polymerization. (a) State two differences between addition polymerization and condensation polymerization. (b) A polyester can be formed from a dicarboxylic acid and a diol monomer. (i) Represent the dicarboxylic acid monomer as \( \text{HOOC}-\Box-\text{COOH} \) and the diol monomer as \( \text{HO}-\bigcirc-\text{OH} \). Draw the structure of the repeating unit of the polyester formed when these two monomers react, showing the ester linkage clearly with all bonds. (ii) State the name of the small molecule eliminated during this condensation reaction. (c) Terylene is a synthetic polyester. State one common household or industrial use of Terylene. (d) Poly(lactic acid), PLA, is a biodegradable polyester. (i) Explain the environmental advantages of biodegradable plastics over non-biodegradable ones. (ii) Lactic acid has the structure \( \text{CH}_3-\text{CH(OH)}-\text{COOH} \). Explain why lactic acid can polymerize with itself to form a polyester, without needing a second monomer.
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Worked solution
(a) Two differences: 1. Addition polymerization involves monomers with C=C double bonds, whereas condensation polymerization involves monomers with functional groups (e.g. -COOH, -OH). 2. Addition polymerization produces only the polymer, whereas condensation polymerization produces the polymer plus a small molecule (like water). (b)(i) Repeating unit structure: \( -[\text{O}-\bigcirc-\text{O}-\text{CO}-\Box-\text{CO}]- \). The diagram must show the ester link \( -\text{O}-\text{C}(=\text{O})- \) clearly with open bonds at both ends of the repeat unit. (b)(ii) Water (\( \text{H}_2\text{O} \)). (c) Clothing, synthetic fibres, sleeping bag fillers, or plastic bottles. (d)(i) Biodegradable plastics are broken down by microbes/bacteria into harmless substances, reducing landfill waste and minimizing harm/entanglement to wildlife. (d)(ii) Lactic acid contains both a carboxylic acid group (\( -\text{COOH} \)) and an alcohol group (\( -\text{OH} \)) within the same molecule. Thus, the acid group of one lactic acid molecule can react with the alcohol group of another lactic acid molecule to form an ester linkage.
Marking scheme
(a) [2 marks]: 1 mark for each valid difference (e.g., types of monomers / functional groups involved, or presence of a small molecule byproduct). (b)(i) [3 marks]: 1 mark for correct ester linkage shown; 1 mark for correct attachment to blocks; 1 mark for open-ended bonds on both sides of the repeating unit. (b)(ii) [1 mark]: Water / H2O. (c) [1 mark]: Clothing / fabric / fibres / sails / bottles. (d)(i) [2 marks]: 1 mark for stating they are decomposed by microorganisms/bacteria; 1 mark for stating they do not accumulate in landfills / reduce litter. (d)(ii) [4 marks]: 1 mark for identifying the carboxylic acid group; 1 mark for identifying the alcohol/hydroxyl group; 1 mark for stating both are present in a single molecule; 1 mark for describing the reaction of the acid group of one monomer with the alcohol group of another to form an ester link.
Paper 61 Alternative to Practical
Answer all questions. Write your answers in the spaces provided on the question paper. Notes for use in qualitative analysis are provided.
4 Question · 40 marks
Question 1 · Practical & Planning
10 marks
A student investigated the rate of reaction between excess calcium carbonate (marble chips) and \(50\text{ cm}^3\) of dilute hydrochloric acid, \(\text{HCl}\), at room temperature.
The apparatus used is shown below (schematic description): A conical flask containing the reactant mixture is sealed with a stopper. A delivery tube connects the stopper of this flask to a gas syringe.
(a) Name the piece of apparatus that should be used to: (i) measure the \(50\text{ cm}^3\) of dilute hydrochloric acid. [1] (ii) collect and measure the volume of carbon dioxide gas produced. [1]
(b) The gas syringe readings at different times during the experiment are shown below. - At \(0\text{ s}\): \(0\text{ cm}^3\) - At \(30\text{ s}\): \(24\text{ cm}^3\) - At \(60\text{ s}\): \(39\text{ cm}^3\) - At \(90\text{ s}\): \(48\text{ cm}^3\) - At \(120\text{ s}\): \(52\text{ cm}^3\) - At \(150\text{ s}\): \(52\text{ cm}^3\)
(i) Explain why the volume of gas in the syringe did not change between \(120\text{ s}\) and \(150\text{ s}\). [1] (ii) Determine which reactant was completely used up by the end of the reaction. Explain your reasoning. [2]
(c) Describe a chemical test to confirm that the gas collected in the syringe is carbon dioxide. test: ......................................................................................................... result: ...................................................................................................... [2]
(d) The student repeated the experiment using the same mass of calcium carbonate and the same volume and concentration of hydrochloric acid, but at a higher temperature. (i) State how the rate of reaction at the higher temperature compares to the rate at room temperature. [1] (ii) Explain, in terms of particles, how increasing the temperature increases the rate of reaction. [2]
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Worked solution
(a) (i) A measuring cylinder is the most appropriate common laboratory apparatus to measure a fixed volume of liquid like \(50\text{ cm}^3\) quickly, though a burette or pipette is also acceptable. (ii) A gas syringe is designed specifically to collect and measure the volume of gas produced.
(b) (i) When the reaction is complete, no more gas is produced, so the volume in the syringe remains constant at \(52\text{ cm}^3\). (ii) Hydrochloric acid was the limiting reactant (completely used up) because the question states that calcium carbonate was in excess.
(c) The standard chemical test for carbon dioxide is bubbling it through limewater (aqueous calcium hydroxide). A positive result is the formation of a white precipitate, which makes the limewater look milky/cloudy.
(d) (i) Increasing temperature increases the rate of a chemical reaction. (ii) At a higher temperature, the reactant particles have more kinetic energy. This leads to more frequent collisions between particles. Additionally, a greater proportion of the colliding particles have energy equal to or exceeding the activation energy, resulting in a higher rate of successful collisions.
Marking scheme
**Part (a)** - (i) [1] Measuring cylinder (accept burette / pipette) - (ii) [1] Gas syringe
**Part (b)** - (i) [1] Reaction has finished / stopped (all acid used up / reactants used up) - (ii) [1] Hydrochloric acid / \(HCl\) - [1] Reason: Calcium carbonate was in excess, so acid must be the limiting reactant / completely used up
**Part (d)** - (i) [1] Rate is faster / higher / increased - (ii) [1] Particles have more kinetic energy / move faster, leading to more frequent collisions - [1] More particles have energy greater than or equal to the activation energy, leading to a higher proportion of successful collisions
Question 2 · Practical & Planning
10 marks
A student investigated a solid mixture, solid **X**, which contains two salts. Solid **X** was dissolved in distilled water to make solution **X**. Portions of solution **X** were tested. The observations are shown in the table below.
| Test | Observation | |---|---| | **Test 1** To the first portion of solution **X**, a few drops of aqueous sodium hydroxide were added. Then, an excess of aqueous sodium hydroxide was added. | White precipitate formed.
Precipitate dissolved in excess to form a colorless solution. | | **Test 2** To the second portion of solution **X**, a few drops of aqueous ammonia were added. Then, an excess of aqueous ammonia was added. | White precipitate formed.
Precipitate remained insoluble in excess. | | **Test 3** To the third portion of solution **X**, dilute nitric acid and aqueous silver nitrate were added. | Cream precipitate formed. | | **Test 4** To the fourth portion of solution **X**, dilute hydrochloric acid and aqueous barium nitrate were added. | No change / colorless solution remained. |
(a) Identify the halide ion present in solid **X**. [1] (b) Identify the cation that gives the results in Test 1 and Test 2. [1] (c) Solid **X** also contains another cation. When a small portion of solid **X** is heated in a dry test-tube with aqueous sodium hydroxide, a gas is evolved. (i) Describe how to test for this gas and state the expected result. test: ......................................................................................................... result: ...................................................................................................... [2] (ii) Identify this second cation. [1] (iii) Identify the gas evolved. [1]
(d) A separate flame test was performed on a sample of solid **X**. (i) Describe how a flame test is carried out. [3] (ii) State the expected flame color if calcium ions were present in solid **X**. [1]
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Worked solution
(a) The formation of a cream precipitate with nitric acid and silver nitrate confirms the presence of bromide ions (\(Br^-\)). (b) A white precipitate that dissolves in excess aqueous sodium hydroxide but is insoluble in excess aqueous ammonia is characteristic of the aluminium ion (\(Al^{3+}\)). (c) (i) Heating an ammonium salt with sodium hydroxide produces ammonia gas. The test for ammonia gas is that it turns damp red litmus paper blue. (ii) The cation that produces ammonia when heated with sodium hydroxide is the ammonium ion (\(NH_4^+\)). (iii) The gas evolved is ammonia (\(NH_3\)). (d) (i) A flame test is performed by cleaning a wire (made of unreactive metal like platinum or nichrome) with concentrated hydrochloric acid, touching it to the solid to be tested, and holding it in the hot, blue zone of a Bunsen burner flame. (ii) Calcium ions produce an orange-red / brick-red color in a flame test.
Marking scheme
**Part (a)** - [1] Bromide (accept \(Br^-\))
**Part (b)** - [1] Aluminium (accept \(Al^{3+}\))
**Part (c)** - (i) [1] Test: damp red litmus paper - [1] Result: turns blue - (ii) [1] Ammonium (accept \(NH_4^+\)) - (iii) [1] Ammonia (accept \(NH_3\))
**Part (d)** - (i) [1] Use a platinum / nichrome wire - [1] Clean the wire with concentrated hydrochloric acid - [1] Place the sample on the wire into a blue / roaring Bunsen burner flame - (ii) [1] Orange-red / brick-red
Question 3 · Practical & Planning
10 marks
A student is provided with a mixture of three solid substances: - Nickel(II) carbonate (insoluble in water, reacts with dilute sulfuric acid to form soluble nickel(II) sulfate) - Barium sulfate (insoluble in water, does not react with dilute sulfuric acid) - Sodium chloride (soluble in water, does not react with dilute sulfuric acid)
Plan an investigation to obtain a pure, dry sample of solid barium sulfate and a pure, dry sample of nickel(II) sulfate-6-water crystals (\(NiSO_4\cdot6H_2O\)) from this mixture.
You are provided with: - The mixture of the three solids - Dilute sulfuric acid - Distilled water - Common laboratory apparatus
Your plan should include: - The steps you would take and the order in which you would perform them - The names of the practical techniques used - How you would obtain the pure, dry solids. [10]
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Worked solution
This is a multi-step separation and preparation of salt process: 1. Separation of soluble and insoluble components: Add distilled water to the mixture and stir. The sodium chloride dissolves, while nickel(II) carbonate and barium sulfate remain insoluble. 2. First filtration: Filter the mixture. The filtrate contains sodium chloride solution (which can be discarded). The residue contains nickel(II) carbonate and barium sulfate. Wash the residue with distilled water to ensure no soluble sodium chloride remains. 3. Chemical separation: Add excess dilute sulfuric acid to the residue. Nickel(II) carbonate reacts: \(NiCO_3 + H_2SO_4 \rightarrow NiSO_4 + H_2O + CO_2\). Barium sulfate does not react and remains insoluble. 4. Second filtration: Filter the mixture. The residue is barium sulfate. Wash it with distilled water to remove any acid or nickel(II) sulfate solution, then dry it (e.g., in an oven or between filter papers) to obtain pure, dry barium sulfate. 5. Crystallization: Heat the filtrate containing nickel(II) sulfate solution in an evaporating basin until the crystallization point is reached (test by seeing if crystals form on a cold glass rod). Let the solution cool slowly to form large crystals of \(NiSO_4\cdot6H_2O\). Filter the crystals, wash them with a small amount of cold distilled water, and leave them to dry.
Marking scheme
**Method of marking: Max 10 marks total from the following points:** - [1] Add distilled water to the mixture and stir / dissolve sodium chloride. - [1] Filter the mixture to separate the insoluble solids (residue) from the sodium chloride solution (filtrate). - [1] Wash the residue (containing nickel(II) carbonate and barium sulfate) with distilled water. - [1] Add dilute sulfuric acid to the residue. - [1] Warm / stir to ensure reaction of nickel(II) carbonate (barium sulfate remains unreacted). - [1] Filter the mixture to separate the barium sulfate residue from the nickel(II) sulfate solution (filtrate). - [1] Wash the barium sulfate residue with distilled water and dry (e.g., in a warm oven or using filter papers). - [1] Heat the filtrate / nickel(II) sulfate solution to the point of crystallization / to evaporate water to form a saturated solution. - [1] Leave the solution to cool slowly to form crystals. - [1] Filter the crystals, wash with a small volume of cold distilled water, and dry (e.g., with filter papers / desiccator; do not accept heating to dryness in oven for hydrated crystals).
Question 4 · practical
10 marks
An eco-friendly solid hand-warmer contains a mixture of iron powder and sodium chloride. When water is added, the iron rusts exothermically in the presence of air. Sodium chloride acts as a catalyst for this process. Plan an investigation to determine how the concentration of sodium chloride solution used affects the maximum temperature change during the reaction. You are provided with: - Iron powder - Sodium chloride solutions of concentrations: 0.5 mol/dm3, 1.0 mol/dm3, 1.5 mol/dm3, 2.0 mol/dm3, 2.5 mol/dm3 - Access to standard laboratory apparatus. Your plan should include: - the apparatus required, - a detailed experimental method, - the variables that must be controlled, - how the results would be processed to draw a conclusion. You may draw a diagram if it helps your explanation.
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Worked solution
Apparatus needed: Polystyrene cup, thermometer, measuring cylinder, mass balance. Method: 1. Weigh a fixed mass of iron powder (e.g., 2.0 g) using a balance and place it into a polystyrene cup. 2. Measure a fixed volume (e.g., 25 cm3) of the 0.5 mol/dm3 sodium chloride solution using a measuring cylinder. 3. Measure and record the initial temperature of the solution using a thermometer. 4. Add the solution to the polystyrene cup containing the iron powder. 5. Stir the mixture and record the maximum temperature reached. 6. Calculate the temperature change by subtracting the initial temperature from the maximum temperature. 7. Repeat the procedure using the same mass of iron and same volume of each of the remaining sodium chloride solutions (1.0, 1.5, 2.0, 2.5 mol/dm3). Variables to control: Keep the mass of iron powder constant, the volume of sodium chloride solution constant, and the starting temperature of the solution constant. Processing results: Plot a graph of the temperature change against the concentration of the sodium chloride solution to determine the relationship.
Marking scheme
Apparatus (Max 2 marks): M1 - Use of a polystyrene cup / insulated container to minimize heat loss (1 mark). M2 - Use of a thermometer AND a measuring cylinder / balance (1 mark). Experimental Method (Max 4 marks): M3 - Measure and record the initial temperature of the sodium chloride solution (1 mark). M4 - Add a known volume of sodium chloride solution to a known mass of iron powder in the cup and stir (1 mark). M5 - Measure and record the highest/maximum temperature reached (1 mark). M6 - Repeat the experiment with all the other concentrations of sodium chloride solution (1 mark). Variables to Control (Max 2 marks): M7 - Keep the mass of iron powder constant (1 mark). M8 - Keep the volume of sodium chloride solution constant / same starting temperature of solution (1 mark). Processing and Conclusion (Max 2 marks): M9 - Calculate the temperature rise for each concentration (1 mark). M10 - Plot a graph of temperature rise against concentration of sodium chloride (1 mark).
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