An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V3) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
Paper 2 Extended Multiple Choice
Answer all 40 multiple-choice questions on the answer sheet. Each correct answer scores 1 mark.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
Under the same conditions of temperature and pressure, which statement about the rate of diffusion of carbon dioxide, \(\text{CO}_2\), and argon, \(\text{Ar}\), is correct?
A.Argon diffuses faster than carbon dioxide because it has a lower relative molecular mass.
B.Argon diffuses slower than carbon dioxide because it is a monatomic noble gas.
C.Carbon dioxide diffuses faster than argon because its molecules contain more atoms.
D.Both gases diffuse at the exact same rate because they are at the same temperature.
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Worked solution
The rate of diffusion of a gas is inversely related to its relative molecular mass (\(M_r\)); lighter molecules diffuse faster. The relative molecular mass of argon (\(\text{Ar}\)) is 40, while that of carbon dioxide (\(\text{CO}_2\)) is \(12 + (16 \times 2) = 44\). Since \(40 < 44\), argon molecules are lighter and will diffuse faster than carbon dioxide.
Marking scheme
Award 1 mark for the correct option A.
Question 2 · multiple-choice
1 marks
What is the total number of atoms in \(0.20\text{ mol}\) of ethanoic acid, \(\text{CH}_3\text{COOH}\)? (The Avogadro constant is \(6.02 \times 10^{23}\text{ /mol}\).)
A.\(1.2 \times 10^{23}\)
B.\(2.4 \times 10^{23}\)
C.\(9.6 \times 10^{23}\)
D.\(4.8 \times 10^{24}\)
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Worked solution
One molecule of ethanoic acid (\(\text{CH}_3\text{COOH}\)) consists of 8 atoms (\(2 \times \text{C}\), \(4 \times \text{H}\), \(2 \times \text{O}\)). Therefore, \(1\text{ mol}\) of \(\text{CH}_3\text{COOH}\) contains \(8\text{ mol}\) of atoms. In \(0.20\text{ mol}\) of ethanoic acid, the number of moles of atoms is \(0.20 \times 8 = 1.6\text{ mol}\). The total number of atoms is \(1.6\text{ mol} \times 6.02 \times 10^{23}\text{ /mol} = 9.632 \times 10^{23}\) atoms (approx \(9.6 \times 10^{23}\)).
Marking scheme
Award 1 mark for the correct option C.
Question 3 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert platinum electrodes. Which row correctly identifies the substances produced at each electrode and the change in the remaining solution?
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride: - At the anode (+), chloride ions (\(\text{Cl}^-\)) are selectively discharged in preference to hydroxide ions (\(\text{OH}^-\)) because they are in high concentration, producing chlorine gas (\(\text{Cl}_2\)). - At the cathode (-), hydrogen ions (\(\text{H}^+\)) are discharged in preference to sodium ions (\(\text{Na}^+\)) because hydrogen is lower in the reactivity series, producing hydrogen gas (\(\text{H}_2\)). - Sodium ions (\(\text{Na}^+\)) and hydroxide ions (\(\text{OH}^-\)) remain in the solution, making it alkaline (sodium hydroxide, \(\text{NaOH}\)).
Marking scheme
Award 1 mark for the correct option C.
Question 4 · multiple-choice
1 marks
Which statement explains why an increase in temperature increases the rate of a chemical reaction?
A.The activation energy of the reaction is lowered.
B.The reactant particles collide more frequently, and a higher proportion of collisions have energy equal to or greater than the activation energy.
C.The reactant particles move more slowly, increasing the chance of successful orientation.
D.The frequency of collisions decreases, but each individual collision transfers a larger amount of energy.
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Worked solution
Increasing the temperature increases the average kinetic energy of the reactant particles, causing them to move faster. This results in more frequent collisions. Crucially, it significantly increases the proportion of colliding particles that possess energy equal to or greater than the activation energy, leading to a much higher frequency of successful collisions.
Marking scheme
Award 1 mark for the correct option B.
Question 5 · multiple-choice
1 marks
The equation for a reversible gaseous reaction is shown.
Which combination of temperature and pressure will produce the highest equilibrium yield of gas C?
A.High temperature and high pressure
B.High temperature and low pressure
C.Low temperature and high pressure
D.Low temperature and low pressure
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Worked solution
To obtain the highest equilibrium yield of product C: 1. Temperature: Since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature will shift the equilibrium position to the right (in the exothermic direction) to oppose the change. 2. Pressure: There are 3 moles of gas on the left (reactant side) and 2 moles of gas on the right (product side). Increasing the pressure will shift the equilibrium position to the right (towards the side with fewer gas moles) to oppose the change. Therefore, low temperature and high pressure will produce the highest equilibrium yield of gas C.
Marking scheme
Award 1 mark for the correct option C.
Question 6 · multiple-choice
1 marks
Propanoic acid reacts with solid calcium carbonate. Which products are formed in this chemical reaction?
A.calcium propanoate and hydrogen only
B.calcium propanoate, carbon dioxide and water
C.calcium propanoate and water only
D.calcium oxide, carbon dioxide and propane
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Worked solution
Carboxylic acids react with metal carbonates to form a salt, carbon dioxide gas, and water. Propanoic acid (\(\text{C}_2\text{H}_5\text{COOH}\)) reacts with calcium carbonate (\(\text{CaCO}_3\)) to produce the salt calcium propanoate (\(\text{(C}_2\text{H}_5\text{COO)}_2\text{Ca}\)), carbon dioxide (\(\text{CO}_2\)), and water (\(\text{H}_2\text{O}\)).
Marking scheme
Award 1 mark for the correct option B.
Question 7 · multiple-choice
1 marks
Four metal oxides are heated separately with carbon powder: - copper(II) oxide - iron(III) oxide - magnesium oxide - sodium oxide
How many of these oxides can be reduced to the corresponding metal by heating with carbon?
A.1
B.2
C.3
D.4
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Worked solution
Only metals that are less reactive than carbon in the reactivity series can be extracted from their oxides by reduction with carbon. Copper and iron are both less reactive than carbon, so copper(II) oxide and iron(III) oxide can be reduced. Sodium and magnesium are more reactive than carbon, so their oxides cannot be reduced by heating with carbon. Thus, exactly 2 of the oxides can be reduced.
Marking scheme
Award 1 mark for the correct option B.
Question 8 · multiple-choice
1 marks
Which synthetic polymer is formed by a condensation polymerisation reaction and contains amide linkages?
A.nylon
B.poly(ethene)
C.Terylene
D.poly(chloroethene)
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Worked solution
Nylon is a polyamide made by the condensation polymerisation of a dicarboxylic acid and a diamine, forming amide linkages (\(\text{-CONH-}\)) with the elimination of water molecules. Terylene is a polyester containing ester linkages. Poly(ethene) and poly(chloroethene) are formed by addition polymerisation.
Marking scheme
Award 1 mark for the correct option A.
Question 9 · multiple-choice
1 marks
Which statement about the rates of diffusion of gaseous sulfur dioxide, \(\text{SO}_2\), carbon dioxide, \(\text{CO}_2\), and carbon monoxide, \(\text{CO}\), is correct?
A.Carbon dioxide diffuses faster than carbon monoxide because its molecules have a larger mass.
B.Carbon dioxide and sulfur dioxide diffuse at the same rate because they are both acidic oxides.
C.Carbon monoxide diffuses faster than sulfur dioxide because its molecules have a smaller molecular mass.
D.Sulfur dioxide diffuses the fastest because it has the largest molecular mass.
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Worked solution
The rate of diffusion of a gas is inversely proportional to its relative molecular mass, \(M_{\text{r}}\).
Since \(\text{CO}\) has the smallest molecular mass, it diffuses the fastest. Since \(\text{SO}_2\) has the largest molecular mass, it diffuses the slowest. Therefore, \(\text{CO}\) diffuses faster than \(\text{SO}_2\) because its molecules have a smaller molecular mass.
Marking scheme
Award 1 mark for the correct option C.
Question 10 · multiple-choice
1 marks
Excess dilute sulfuric acid is added to \(5.4\text{ g}\) of aluminium powder.
Which volume of hydrogen gas, measured at room temperature and pressure (r.t.p.), is produced?
[Relative atomic mass, \(A_{\text{r}}\): \(\text{Al} = 27\); the volume of one mole of any gas is \(24\text{ dm}^3\) at r.t.p.]
A.2.4 dm³
B.4.8 dm³
C.7.2 dm³
D.14.4 dm³
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Worked solution
Step 1: Calculate the number of moles of aluminium. \[\text{moles of Al} = \frac{5.4\text{ g}}{27\text{ g/mol}} = 0.20\text{ mol}\]
Step 2: Use the stoichiometric ratio from the equation to find the moles of hydrogen gas. From the equation, \(2\text{ mol}\) of \(\text{Al}\) produces \(3\text{ mol}\) of \(\text{H}_2\). \[\text{moles of H}_2 = 0.20\text{ mol} \times \frac{3}{2} = 0.30\text{ mol}\]
Step 3: Calculate the volume of hydrogen gas at r.t.p. \[\text{Volume} = 0.30\text{ mol} \times 24\text{ dm}^3/\text{mol} = 7.2\text{ dm}^3\]
Marking scheme
Award 1 mark for the correct option C.
Question 11 · multiple-choice
1 marks
Concentrated aqueous copper(II) chloride is electrolysed using graphite electrodes. Which row describes the correct observations at each electrode?
A.Positive electrode: bubbles of a green gas; Negative electrode: pink-brown solid formed
B.Positive electrode: bubbles of a colourless gas; Negative electrode: bubbles of a colourless gas
C.Positive electrode: pink-brown solid formed; Negative electrode: bubbles of a green gas
D.Positive electrode: bubbles of a colourless gas; Negative electrode: pink-brown solid formed
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Worked solution
During the electrolysis of concentrated aqueous copper(II) chloride: - At the positive electrode (anode), halide ions (\(\text{Cl}^-\)) are discharged preferentially over hydroxide ions (\(\text{OH}^-\)) because the solution is concentrated. This produces chlorine gas, which is observed as bubbles of a green-yellow gas. - At the negative electrode (cathode), copper ions (\(\text{Cu}^{2+}\)) are discharged preferentially over hydrogen ions (\(\text{H}^+\)) because copper is lower in the reactivity series. This produces copper metal, observed as a pink-brown solid deposit.
Marking scheme
Award 1 mark for the correct option A.
Question 12 · multiple-choice
1 marks
The chemical equation for the synthesis of ammonia is shown.
| Bond | Bond energy / kJ/mol | | :--- | :--- | | \(\text{N}\equiv\text{N}\) | 945 | | \(\text{H}-\text{H}\) | 436 | | \(\text{N}-\text{H}\) | 390 |
What is the energy change for this reaction?
A.−87 kJ/mol
B.+87 kJ/mol
C.−109 kJ/mol
D.+109 kJ/mol
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Worked solution
Step 1: Calculate the energy required to break the reactant bonds (endothermic process). - Breaking \(1\) mole of \(\text{N}\equiv\text{N}\) bonds: \(1 \times 945 = 945\text{ kJ}\) - Breaking \(3\) moles of \(\text{H}-\text{H}\) bonds: \(3 \times 436 = 1308\text{ kJ}\) Total energy in = \(945 + 1308 = 2253\text{ kJ}\)
Step 2: Calculate the energy released when product bonds are formed (exothermic process). - Forming \(6\) moles of \(\text{N}-\text{H}\) bonds (since each of the \(2\) molecules of \(\text{NH}_3\) has \(3\) \(\text{N}-\text{H}\) bonds): Total energy out = \(6 \times 390 = 2340\text{ kJ}\)
Step 3: Calculate the overall energy change (\(\Delta H\)). \(\Delta H = \text{energy in} - \text{energy out} = 2253 - 2340 = -87\text{ kJ/mol}\)
Marking scheme
Award 1 mark for the correct option A.
Question 13 · multiple-choice
1 marks
The reaction for the catalytic oxidation of sulfur dioxide is reversible and exothermic in the forward direction.
Which set of conditions produces the highest yield of sulfur trioxide, \(\text{SO}_3\), at equilibrium?
A.High temperature and high pressure
B.High temperature and low pressure
C.Low temperature and high pressure
D.Low temperature and low pressure
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Worked solution
- Temperature: The forward reaction is exothermic. According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium in the direction of the exothermic reaction (to the right), increasing the yield of \(\text{SO}_3\). - Pressure: There are \(3\) moles of gaseous reactants on the left-hand side and \(2\) moles of gaseous products on the right-hand side. Increasing the pressure shifts the equilibrium towards the side with fewer gas moles (to the right), increasing the yield of \(\text{SO}_3\).
Therefore, a low temperature and high pressure will produce the highest yield of sulfur trioxide.
Marking scheme
Award 1 mark for the correct option C.
Question 14 · multiple-choice
1 marks
Aqueous halogens are added to separate solutions of potassium halides as shown below:
1. Aqueous chlorine is added to aqueous potassium bromide. 2. Aqueous bromine is added to aqueous potassium iodide.
Which row correctly describes the colour changes observed in these reactions?
A.Reaction 1: colourless to orange; Reaction 2: orange to brown
B.Reaction 1: colourless to orange; Reaction 2: no change
C.Reaction 1: no change; Reaction 2: orange to brown
D.Reaction 1: no change; Reaction 2: no change
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Worked solution
- Reaction 1: Chlorine is more reactive than bromine, so it displaces bromide ions from potassium bromide. Bromine gas/liquid dissolved in water is produced, turning the colourless solution orange: \(\text{Cl}_2\text{(aq)} + 2\text{KBr(aq)} \rightarrow 2\text{KCl(aq)} + \text{Br}_2\text{(aq)}\)
- Reaction 2: Bromine is more reactive than iodine, so it displaces iodide ions from potassium iodide. Aqueous iodine is produced, turning the orange bromine water to a brown solution: \(\text{Br}_2\text{(aq)} + 2\text{KI(aq)} \rightarrow 2\text{KBr(aq)} + \text{I}_2\text{(aq)}\)
Thus, both mixtures show a distinct color change as described in row A.
Marking scheme
Award 1 mark for the correct option A.
Question 15 · multiple-choice
1 marks
Propanoic acid reacts with methanol in the presence of an acid catalyst to form an organic compound and water.
What is the name and structural formula of the organic compound formed?
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Worked solution
Carboxylic acids react with alcohols to form esters and water in a condensation reaction. - Propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) reacts with methanol (\(\text{CH}_3\text{OH}\)). - The ester formed is named by taking the prefix from the alcohol ('methyl') and the suffix from the carboxylic acid ('propanoate'), giving 'methyl propanoate'. - The structural formula of methyl propanoate is \(\text{CH}_3\text{CH}_2\text{COOCH}_3\).
Marking scheme
Award 1 mark for the correct option B.
Question 16 · multiple-choice
1 marks
An unknown salt X is dissolved in water to form a solution. Two tests are performed on separate portions of this solution:
- When aqueous sodium hydroxide is added, a green precipitate is formed that is insoluble in excess. - When dilute nitric acid followed by aqueous barium nitrate is added, a white precipitate is formed.
What is the identity of salt X?
A.chromium(III) sulfate
B.iron(II) sulfate
C.iron(II) chloride
D.iron(III) sulfate
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Worked solution
- Test 1 (with sodium hydroxide): The formation of a green precipitate that is insoluble in excess sodium hydroxide is the characteristic test for iron(II) ions, \(\text{Fe}^{2+}\). (Note: Chromium(III) also forms a green precipitate, but it dissolves in excess sodium hydroxide to form a green solution). - Test 2 (with acidified barium nitrate): The formation of a white precipitate is the characteristic test for sulfate ions, \(\text{SO}_4^{2-}\).
Therefore, salt X is iron(II) sulfate, \(\text{FeSO}_4\).
Marking scheme
Award 1 mark for the correct option B.
Question 17 · multiple-choice
1 marks
What is the total number of shared pairs of electrons in a 0.1 mol sample of methane, \(\text{CH}_4\)? (Avogadro constant, \(L = 6.0 \times 10^{23}\ \text{mol}^{-1}\))
A.\(6.0 \times 10^{22}\)
B.\(2.4 \times 10^{23}\)
C.\(6.0 \times 10^{23}\)
D.\(2.4 \times 10^{24}\)
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Worked solution
Each methane molecule, \(\text{CH}_4\), contains 4 single covalent bonds, which correspond to 4 shared pairs of electrons. The number of molecules in a 0.1 mol sample is calculated as: \(0.1 \text{ mol} \times 6.0 \times 10^{23} \text{ molecules/mol} = 6.0 \times 10^{22} \text{ molecules}\). The total number of shared pairs of electrons is: \(4 \times 6.0 \times 10^{22} = 2.4 \times 10^{23}\).
Marking scheme
Award 1 mark for the correct option B. Reject all other options.
Question 18 · multiple-choice
1 marks
Which pair of compounds are structural isomers of each other?
A.\(\text{CH}_3\text{CH}_2\text{COOCH}_3\) and \(\text{CH}_3\text{COOCH}_2\text{CH}_3\)
B.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) and \(\text{CH}_3\text{CH}_2\text{CHO}\)
C.\(\text{CH}_3\text{CH}_2\text{COOH}\) and \(\text{CH}_3\text{CH}_2\text{COOCH}_3\)
D.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\) and \(\text{CH}_3\text{CH}=\text{CHCH}_3\)
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Worked solution
Structural isomers share the same molecular formula but have different structural arrangements. Both methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)) and ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)) have the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\), making them structural isomers.
Marking scheme
Award 1 mark for the correct option A. Reject all other options.
Question 19 · multiple-choice
1 marks
An aqueous solution of sodium sulfate, \(\text{Na}_2\text{SO}_4\), is electrolysed using inert platinum electrodes. Which row correctly identifies the products formed at the positive electrode (anode) and the negative electrode (cathode)?
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Worked solution
At the positive electrode (anode), \(\text{OH}^-\text{(aq)}\) ions from water are discharged in preference to \(\text{SO}_4^{2-}\text{(aq)}\) to form oxygen gas (\(\text{O}_2\)). At the negative electrode (cathode), \(\text{H}^+\text{(aq)}\) ions are discharged in preference to \(\text{Na}^+\text{(aq)}\) to form hydrogen gas (\(\text{H}_2\)).
Marking scheme
Award 1 mark for the correct option A. Reject all other options.
Question 20 · multiple-choice
1 marks
The reaction between calcium carbonate and dilute hydrochloric acid is investigated. Why does an increase in the temperature of the acid increase the rate of reaction?
1. The reacting particles collide more frequently. 2. A greater proportion of the colliding particles have energy equal to or greater than the activation energy. 3. The activation energy of the reaction decreases.
Which statements are correct?
A.1 and 2 only
B.1 and 3 only
C.2 and 3 only
D.1, 2 and 3
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Worked solution
Increasing temperature increases the average kinetic energy of the particles, causing them to move faster and collide more frequently (Statement 1). It also significantly increases the proportion of colliding particles that possess energy equal to or greater than the activation energy (Statement 2). The activation energy itself remains unchanged (Statement 3 is incorrect, as activation energy is only affected by adding a catalyst).
Marking scheme
Award 1 mark for the correct option A. Reject all other options.
Question 21 · multiple-choice
1 marks
The reaction in the Contact process is reversible:
Which set of conditions will produce the greatest yield of sulfur trioxide, \(\text{SO}_3\), at equilibrium?
A.temperature: high; pressure: high
B.temperature: high; pressure: low
C.temperature: low; pressure: high
D.temperature: low; pressure: low
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Worked solution
Because the forward reaction is exothermic, a lower temperature shifts the equilibrium position to the right, favoring the forward reaction and increasing the yield of sulfur trioxide. Since there are fewer moles of gas on the product side (2 moles) than on the reactant side (3 moles), a higher pressure also shifts the equilibrium position to the right. Therefore, a combination of low temperature and high pressure maximizes the yield.
Marking scheme
Award 1 mark for the correct option C. Reject all other options.
Question 22 · multiple-choice
1 marks
Which method is used to prepare a pure, dry sample of the insoluble salt, lead(II) sulfate?
A.Titration of aqueous lead(II) nitrate with dilute sulfuric acid, followed by evaporation.
B.Precipitation by mixing aqueous lead(II) nitrate and dilute sulfuric acid, followed by filtration, washing the residue with distilled water, and drying.
C.Reacting lead metal with dilute sulfuric acid, followed by filtration and crystallisation.
D.Heating lead(II) oxide with dilute sulfuric acid, followed by filtration and evaporation.
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Worked solution
Lead(II) sulfate is an insoluble salt. The standard preparation for an insoluble salt is precipitation, where two soluble solutions (aqueous lead(II) nitrate and dilute sulfuric acid) are mixed. The insoluble lead(II) sulfate precipitates out and is separated by filtration, washed with distilled water to remove residues of soluble reactants, and dried.
Marking scheme
Award 1 mark for the correct option B. Reject all other options.
Question 23 · multiple-choice
1 marks
Hydrogen reacts with chlorine to form hydrogen chloride as shown:
| Bond | Bond energy / kJ/mol | | :---: | :---: | | H-H | 436 | | Cl-Cl | 242 | | H-Cl | 431 |
What is the enthalpy change, \(\Delta H\), for this reaction?
A.\(-184\text{ kJ/mol}\)
B.\(+184\text{ kJ/mol}\)
C.\(-247\text{ kJ/mol}\)
D.\(+247\text{ kJ/mol}\)
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Worked solution
Energy required to break bonds (reactants) = \(436\text{ (H-H)} + 242\text{ (Cl-Cl)} = 678\text{ kJ/mol}\). Energy released when new bonds are formed (products) = \(2 \times 431\text{ (H-Cl)} = 862\text{ kJ/mol}\). Enthalpy change, \(\Delta H = \text{Energy absorbed} - \text{Energy released} = 678 - 862 = -184\text{ kJ/mol}\).
Marking scheme
Award 1 mark for the correct option A. Reject all other options.
Question 24 · multiple-choice
1 marks
Which row correctly describes the trends in density and reactivity of the Halogens (Group VII) as the group is descended?
A.density: decreases; reactivity: increases
B.density: decreases; reactivity: decreases
C.density: increases; reactivity: increases
D.density: increases; reactivity: decreases
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Worked solution
As Group VII is descended, the physical state shifts from gas to liquid to solid, reflecting an increase in density. Reactivity decreases down the group because the atoms get larger and the outer shell is further away from the attractive pull of the nucleus, making it harder to gain an electron.
Marking scheme
Award 1 mark for the correct option D. Reject all other options.
Question 25 · multiple-choice
1 marks
Four gases, \( G_1, G_2, G_3, \) and \( G_4 \), have different relative molecular masses. A student measures the time taken for equal volumes of each gas to diffuse through a small hole under identical conditions.
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Worked solution
According to Graham's law, the rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass (\( M_r \)). A faster rate of diffusion corresponds to a shorter time taken to diffuse. Therefore, gases with lower \( M_r \) diffuse faster and take less time. The order of increasing time to diffuse corresponds directly to the order of increasing relative molecular mass (\( M_r \)):
Therefore, the correct sequence is \( G_3 \rightarrow G_4 \rightarrow G_1 \rightarrow G_2 \).
Marking scheme
1 mark for identifying that lighter gases diffuse faster (and thus take less time to diffuse), and correctly ordering them by increasing relative molecular mass.
Question 26 · multiple-choice
1 marks
Which statement about two isotopes of the same element is correct?
A.They have different chemical properties because they have different numbers of neutrons.
B.They have the same chemical properties because they have the same electronic configuration.
C.They have different physical properties because they have different numbers of protons.
D.They have the same physical properties because they have the same relative atomic mass.
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Worked solution
Isotopes are atoms of the same element with the same number of protons and electrons, but different numbers of neutrons. Because chemical reactions are determined by the electronic configuration (specifically the number of valence electrons), isotopes share identical chemical properties. Physical properties (such as density and mass) differ because of the different numbers of neutrons.
Marking scheme
1 mark for identifying the correct relationship between isotope structures and their chemical properties.
Question 27 · multiple-choice
1 marks
How many ions are present in \( 0.05\text{ mol} \) of aluminium sulfate, \( \text{Al}_2(\text{SO}_4)_3 \)?
A.\( 3.0 \times 10^{22} \)
B.\( 9.0 \times 10^{22} \)
C.\( 1.5 \times 10^{23} \)
D.\( 3.0 \times 10^{23} \)
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Worked solution
One formula unit of aluminium sulfate, \( \text{Al}_2(\text{SO}_4)_3 \), contains 2 aluminium ions (\( \text{Al}^{3+} \)) and 3 sulfate ions (\( \text{SO}_4^{2-} \)), giving a total of 5 ions.
Total moles of ions in \( 0.05\text{ mol} \) of \( \text{Al}_2(\text{SO}_4)_3 \): \( 0.05\text{ mol} \times 5 = 0.25\text{ mol} \) of ions.
Using Avogadro's constant (\( 6.02 \times 10^{23}\text{ mol}^{-1} \)): Number of ions = \( 0.25\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 1.505 \times 10^{23} \) ions.
Marking scheme
1 mark for calculating the total moles of ions (0.25 mol) and multiplying by Avogadro's constant to get \( 1.5 \times 10^{23} \).
Question 28 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert platinum electrodes. Which row correctly describes what is produced at each electrode and the change in the electrolyte?
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride (brine): - At the anode (+), halide ions (\( \text{Cl}^- \)) are in high concentration and are selectively discharged over hydroxide ions (\( \text{OH}^- \)) to form chlorine gas. - At the cathode (-), hydrogen ions (\( \text{H}^+ \)) are discharged rather than sodium ions (\( \text{Na}^+ \)) because hydrogen is lower in the reactivity series, producing hydrogen gas. - Hydroxide ions (\( \text{OH}^- \)) and sodium ions (\( \text{Na}^+ \)) remain in the solution, forming sodium hydroxide, which is alkaline, so the pH of the electrolyte increases.
Marking scheme
1 mark for identifying chlorine at the anode, hydrogen at the cathode, and a pH increase in the electrolyte (Row A).
Question 29 · multiple-choice
1 marks
The reaction between hydrazine (\( \text{N}_2\text{H}_4 \)) and oxygen (\( \text{O}_2 \)) is represented below:
What is the energy change (\( \Delta H \)) for the reaction?
A.-569 kJ/mol
B.+569 kJ/mol
C.-1129 kJ/mol
D.+1129 kJ/mol
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Worked solution
First, calculate the energy required to break all the reactant bonds: - Bonds broken in \( \text{N}_2\text{H}_4 \): \( 1 \times (\text{N}-\text{N}) + 4 \times (\text{N}-\text{H}) = 160 + 4(390) = 1720 \text{ kJ/mol} \). - Bonds broken in \( \text{O}_2 \): \( 1 \times (\text{O}=\text{O}) = 496 \text{ kJ/mol} \). - Total energy input (breaking bonds) = \( 1720 + 496 = 2216 \text{ kJ/mol} \).
Next, calculate the energy released when new bonds in products are formed: - Bonds formed in \( \text{N}_2 \): \( 1 \times (\text{N}\equiv\text{N}) = 945 \text{ kJ/mol} \). - Bonds formed in \( 2\text{H}_2\text{O} \): \( 4 \times (\text{O}-\text{H}) = 4(460) = 1840 \text{ kJ/mol} \). - Total energy output (forming bonds) = \( 945 + 1840 = 2785 \text{ kJ/mol} \).
1 mark for the correct calculation of bond breaking energy (2216 kJ/mol) and bond making energy (2785 kJ/mol), yielding the correct change of -569 kJ/mol.
Question 30 · multiple-choice
1 marks
Which statement about the effect of temperature and concentration on the rate of a chemical reaction is correct?
A.Increasing the temperature increases the activation energy of the reaction.
B.Increasing the concentration increases the rate because particles have more kinetic energy.
C.Increasing the temperature increases the rate because a larger fraction of collisions have energy greater than the activation energy.
D.Increasing the concentration decreases the rate because particles are closer together, reducing collision frequency.
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Worked solution
According to collision theory: - Increasing the temperature increases the average kinetic energy of the particles. Consequently, a much larger fraction of colliding particles have energy equal to or greater than the activation energy (\( E_a \)). This significantly increases the frequency of successful collisions. - Increasing concentration increases the number of particles per unit volume, which increases the frequency of collisions but does *not* increase the kinetic energy of the individual particles. - Activation energy is a constant for a given reaction (unless a catalyst is used).
Marking scheme
1 mark for choosing the correct explanation of temperature effects on the rate of reaction based on the fraction of successful collisions.
Question 31 · multiple-choice
1 marks
In the extraction of iron in the blast furnace, limestone is added. Which row correctly describes the function of limestone and the nature of the reaction that forms slag?
| Row | Function of limestone | Nature of slag formation reaction | | :---: | --- | --- | | **A** | To reduce iron(III) oxide | Acid-base reaction | | **B** | To remove silicon(IV) oxide impurities | Thermal decomposition only | | **|C|** | To remove silicon(IV) oxide impurities | Acid-base reaction | | **D** | To act as a fuel to heat the furnace | Redox reaction |
A.Row A
B.Row B
C.Row C
D.Row D
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Worked solution
Limestone (\( \text{CaCO}_3 \)) is added to remove acidic silicon(IV) oxide (\( \text{SiO}_2 \)) impurities present in the iron ore. First, limestone thermally decomposes to form calcium oxide (\( \text{CaO} \)), a basic oxide. Then, \( \text{CaO} \) reacts with \( \text{SiO}_2 \) to produce slag (\( \text{CaSiO}_3 \)). Because this reaction occurs between a basic oxide and an acidic oxide, it is an acid-base (neutralisation) reaction.
Marking scheme
1 mark for identifying both the correct function of limestone and the acid-base nature of the slag formation reaction (Row C).
Question 32 · multiple-choice
1 marks
Propanoic acid reacts with sodium carbonate. What are the products of this reaction?
A.sodium propanoate and water only
B.sodium propanoate, carbon dioxide and water
C.sodium propyl oxide, carbon dioxide and water
D.propane, sodium hydroxide and carbon dioxide
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Worked solution
Propanoic acid is a carboxylic acid. Carboxylic acids react with metal carbonates to form a salt, carbon dioxide gas, and water:
The products are sodium propanoate (the salt), carbon dioxide, and water.
Marking scheme
1 mark for identifying that the reaction between propanoic acid and sodium carbonate yields sodium propanoate, carbon dioxide, and water.
Question 33 · multiple-choice
1 marks
Two gas syringes, one containing gas P and the other containing gas Q, are placed in a water bath at \(40\ ^\circ\text{C}\). Gas P has a relative molecular mass, \(M_\text{r}\), of 44. Gas Q has a relative molecular mass, \(M_\text{r}\), of 64. Which statement about the diffusion of these two gases is correct?
A.Gas P diffuses slower than gas Q because it has a lower relative molecular mass.
B.Gas P diffuses faster than gas Q because it has a lower relative molecular mass.
C.Both gases diffuse at the same rate because they are at the same temperature.
D.Increasing the temperature to \(60\ ^\circ\text{C}\) decreases the rate of diffusion of both gases.
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Worked solution
According to Graham's law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass (density). Lighter molecules (lower \(M_\text{r}\)) move faster on average at a given temperature, and therefore diffuse more rapidly. Gas P (\(M_\text{r} = 44\)) has a lower relative molecular mass than gas Q (\(M_\text{r} = 64\)), meaning P will diffuse faster than Q.
Marking scheme
Award 1 mark for the correct option B.
Question 34 · multiple-choice
1 marks
What is the total number of atoms in \(0.20\text{ mol}\) of nitrogen gas, \(N_2\)? [Avogadro constant, \(L = 6.02 \times 10^{23}\text{ per mole}\)]
A.\(1.20 \times 10^{23}\)
B.\(2.41 \times 10^{23}\)
C.\(1.20 \times 10^{24}\)
D.\(2.41 \times 10^{24}\)
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Worked solution
First, calculate the number of nitrogen molecules: \(0.20\text{ mol} \times 6.02 \times 10^{23}\text{ molecules/mol} = 1.204 \times 10^{23}\text{ molecules}\). Since nitrogen gas is diatomic, each molecule contains 2 atoms. Total number of atoms = \(1.204 \times 10^{23} \times 2 = 2.41 \times 10^{23}\).
Marking scheme
Award 1 mark for the correct option B.
Question 35 · multiple-choice
1 marks
Concentrated aqueous copper(II) chloride is electrolysed using platinum electrodes. Which row correctly describes the product at each electrode?
A.Anode: hydrogen gas; Cathode: copper metal
B.Anode: chlorine gas; Cathode: copper metal
C.Anode: oxygen gas; Cathode: hydrogen gas
D.Anode: chlorine gas; Cathode: hydrogen gas
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Worked solution
At the cathode (negative electrode), copper ions (\(\text{Cu}^{2+}\)) are discharged in preference to hydrogen ions because copper is lower in the reactivity series, forming copper metal. At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions because it is a concentrated halide solution, forming chlorine gas.
Marking scheme
Award 1 mark for the correct option B.
Question 36 · multiple-choice
1 marks
An ester has the structural formula \(\text{CH}_3\text{COOCH}_2\text{CH}_2\text{CH}_3\). Which alcohol and carboxylic acid are used to manufacture this ester?
A.propan-1-ol and ethanoic acid
B.ethanol and propanoic acid
C.methanol and butanoic acid
D.propan-1-ol and methanoic acid
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Worked solution
An ester is formed from a condensation reaction between an alcohol and a carboxylic acid. The prefix of the ester name comes from the alcohol (propyl group, \(-\text{CH}_2\text{CH}_2\text{CH}_3\), from propan-1-ol), and the suffix comes from the carboxylic acid (ethanoate group, \(\text{CH}_3\text{COO}-\), from ethanoic acid).
Marking scheme
Award 1 mark for the correct option A.
Question 37 · multiple-choice
1 marks
Hydrogen reacts with chlorine according to the equation: \(\text{H}_2(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow 2\text{HCl}(\text{g})\). The reaction is exothermic with an energy change of \(-184\text{ kJ/mol}\). Some bond energies are shown in the table: - \(\text{H}-\text{H}\) bond energy: \(+436\text{ kJ/mol}\) - \(\text{Cl}-\text{Cl}\) bond energy: \(+242\text{ kJ/mol}\) What is the bond energy of the \(\text{H}-\text{Cl}\) bond?
A.\(+247\text{ kJ/mol}\)
B.\(+431\text{ kJ/mol}\)
C.\(+862\text{ kJ/mol}\)
D.\(+494\text{ kJ/mol}\)
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An excess of zinc pieces is reacted with \(50\text{ cm}^3\) of \(1.0\text{ mol/dm}^3\) hydrochloric acid. The volume of hydrogen gas collected is recorded. The experiment is repeated using \(25\text{ cm}^3\) of \(2.0\text{ mol/dm}^3\) hydrochloric acid with excess zinc pieces, with all other conditions kept the same. How do the initial rate of reaction and the final volume of hydrogen gas compare to the first experiment?
A.Initial rate: faster; Final volume of gas: unchanged
B.Initial rate: faster; Final volume of gas: halved
C.Initial rate: unchanged; Final volume of gas: unchanged
D.Initial rate: slower; Final volume of gas: doubled
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Worked solution
First, compare the moles of reactant: \(0.050\text{ dm}^3 \times 1.0\text{ mol/dm}^3 = 0.050\text{ mol HCl}\) in Exp 1, and \(0.025\text{ dm}^3 \times 2.0\text{ mol/dm}^3 = 0.050\text{ mol HCl}\) in Exp 2. Since the moles of the limiting reactant (HCl) are the same, the final volume of gas produced is unchanged. Because the concentration of hydrochloric acid in the second experiment is higher (\(2.0\text{ mol/dm}^3\) vs \(1.0\text{ mol/dm}^3\)), the initial rate of reaction is faster.
Marking scheme
Award 1 mark for the correct option A.
Question 39 · multiple-choice
1 marks
Which statement about weak acids, compared to strong acids of the same concentration, is correct?
A.Weak acids fully dissociate into ions in aqueous solution.
B.Solutions of weak acids have a lower pH than strong acids.
C.Weak acids react more slowly with magnesium ribbon.
D.Weak acids do not conduct electricity at all.
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Worked solution
Weak acids are only partially ionized in aqueous solutions, resulting in a lower concentration of hydrogen ions (\(\text{H}^+\)) than strong acids of the same concentration. Because concentration of \(\text{H}^+\), which acts as a reactant, is lower, the rate of reaction with metals like magnesium is slower.
Marking scheme
Award 1 mark for the correct option C.
Question 40 · multiple-choice
1 marks
Which row correctly describes the trend in properties of the Group VII halogens as the group is descended from fluorine to iodine?
A.Color becomes lighter; reactivity increases
B.Color becomes darker; melting point increases
C.Density decreases; melting point decreases
D.Color becomes darker; reactivity increases
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Worked solution
As Group VII is descended: the colors of the halogens become darker (from pale yellow to green, red-brown, and then grey-black), the melting/boiling points increase due to stronger intermolecular forces, and density increases. Reactivity decreases down the group.
Marking scheme
Award 1 mark for the correct option B.
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8 Question · 80 marks
Question 1 · Structured Theory & Equation Completion
10 marks
Anhydrous zinc carbonate decomposes when heated strongly to produce zinc oxide and carbon dioxide gas according to the equation:
\(ZnCO_3(s) \rightarrow ZnO(s) + CO_2(g)\)
(a) State the colour change observed in the solid during this thermal decomposition. [2]
(b) Calculate the mass of zinc oxide formed when 5.00 g of anhydrous zinc carbonate is completely decomposed. [3]
(c) Calculate the volume of carbon dioxide gas, in \(dm^3\), produced at room temperature and pressure (r.t.p.) from the decomposition of 5.00 g of zinc carbonate. [3]
(d) State the chemical test for carbon dioxide and the positive result. [2]
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Worked solution
(a) Zinc carbonate is a white solid. When heated, it decomposes to zinc oxide, which is yellow when hot and white when cold. Thus, the solid turns yellow during heating and returns to white upon cooling.
(b) First, calculate the relative formula mass (\(M_r\)) of \(ZnCO_3\): \(M_r(ZnCO_3) = 65 + 12 + (3 \times 16) = 125\)
Moles of \(ZnCO_3 = \frac{5.00}{125} = 0.0400\) mol
From the equation, the mole ratio of \(ZnCO_3\) to \(ZnO\) is 1:1, so moles of \(ZnO = 0.0400\) mol.
Calculate the relative formula mass of \(ZnO\): \(M_r(ZnO) = 65 + 16 = 81\)
Mass of \(ZnO = 0.0400 \times 81 = 3.24\) g.
(c) From the equation, the mole ratio of \(ZnCO_3\) to \(CO_2\) is 1:1, so moles of \(CO_2 = 0.0400\) mol.
Volume of \(CO_2\) at r.t.p. = \(0.0400 \times 24 = 0.96\) \(dm^3\).
(d) Carbon dioxide gas is tested by bubbling it through limewater (aqueous calcium hydroxide). A positive result is the limewater turning milky or cloudy due to the formation of a white precipitate of calcium carbonate.
Marking scheme
(a) [2 marks total]: - White to yellow when hot [1] - Yellow to white on cooling / yellow [1]
(b) [3 marks total]: - Calculating moles of \(ZnCO_3 = 0.0400\) mol [1] - Using 1:1 ratio to find moles of \(ZnO = 0.0400\) mol [1] - Mass of \(ZnO = 3.24\) g [1] (allow ecf from incorrect moles)
(c) [3 marks total]: - Moles of \(CO_2 = 0.0400\) mol [1] - Multiplying by 24 [1] - Volume of \(CO_2 = 0.96\) \(dm^3\) [1] (allow 960 \(cm^3\) if unit is specified, allow ecf)
(d) [2 marks total]: - Bubble gas into limewater [1] - Turns milky / cloudy / white precipitate [1]
Question 2 · Structured Theory & Equation Completion
10 marks
Propan-1-ol can be oxidised to form propanoic acid.
(a) State the name of a suitable oxidising agent used for this reaction and the colour change observed. [2]
(b) Draw the fully displayed formula of propanoic acid, showing all atoms and all bonds. [2]
(c) Propanoic acid reacts with methanol in the presence of an acid catalyst to form an ester.
(i) State the name of the acid catalyst used. [1]
(ii) Name the ester formed and write its structural formula. [2]
(iii) Draw the displayed formula of the ester linkage. [1]
(d) State two characteristics of a homologous series. [2]
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Worked solution
(a) The oxidation of alcohols to carboxylic acids is carried out using strong oxidising agents. Suitable agents include: 1. Acidified potassium manganate(VII), which turns from purple to colourless. 2. Acidified potassium dichromate(VI), which turns from orange to green.
(b) Propanoic acid is a 3-carbon carboxylic acid. Its displayed formula must show every single atom (C, H, and O) and every single bond, including the C-H, C-C, C=O, C-O, and O-H bonds:
H H O | | // H - C - C - C | | H H O - H
(c)(i) The catalyst for esterification is concentrated sulfuric acid. (ii) The reaction between propanoic acid and methanol forms the ester methyl propanoate. Its structural formula is \(CH_3CH_2COOCH_3\) or \(C_2H_5COOCH_3\). (iii) The ester linkage consists of a carbon atom double-bonded to an oxygen atom, and single-bonded to another oxygen atom which is connected to a carbon chain: -C(=O)-O-
(d) A homologous series is a family of similar organic compounds. Key characteristics include sharing the same general formula, having similar chemical properties, having the same functional group, and showing a gradual trend in physical properties (consecutive members differ by a \(-\text{CH}_2-\) unit).
Marking scheme
(a) [2 marks total]: - Acidified potassium manganate(VII) / acidified potassium dichromate(VI) [1] - Purple to colourless / orange to green [1]
(b) [2 marks total]: - 3-carbon chain with single C-C bonds and carboxylic acid group shown [1] - All atoms and all bonds shown correctly, including O-H bond [1]
(c) [4 marks total]: - (i) Concentrated sulfuric acid (reject: dilute sulfuric acid) [1] - (ii) Methyl propanoate [1] - (ii) Structural formula: \(CH_3CH_2COOCH_3\) / \(C_2H_5COOCH_3\) [1] - (iii) Displayed ester linkage showing C=O and C-O single bond correctly [1]
(d) [2 marks total]: - Any two from: same general formula [1], similar chemical properties [1], same functional group [1], consecutive members differ by a \(CH_2\) group [1].
Question 3 · Structured Theory & Equation Completion
10 marks
A student electrolyses aqueous copper(II) sulfate using inert carbon (graphite) electrodes.
(a) State the products formed at:
(i) the negative electrode (cathode) [1]
(ii) the positive electrode (anode). [1]
(b) Write an ionic half-equation, including state symbols, for the reaction occurring at the positive electrode. [2]
(c) State and explain any changes observed in the electrolyte solution during this electrolysis. [2]
(d) The student repeats the electrolysis, but replaces the carbon electrodes with copper electrodes.
(i) Describe the observations at both the anode and the cathode. [2]
(ii) Explain why the intensity of the blue colour of the electrolyte remains constant in this second experiment. [2]
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Worked solution
(a)(i) At the negative electrode (cathode), copper(II) ions are discharged to form copper metal (solid): \(Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)\). (ii) At the positive electrode (anode), hydroxide ions from water are preferentially discharged to form oxygen gas: \(4OH^-(aq) \rightarrow O_2(g) + 2H_2O(l) + 4e^-\).
(b) The ionic half-equation for the anode reaction must include the correct species, balancing, and state symbols: \(4OH^-(aq) \rightarrow O_2(g) + 2H_2O(l) + 4e^-\).
(c) The blue colour of the aqueous copper(II) sulfate solution fades and eventually becomes colourless. This is because \(Cu^{2+}\) ions (which give the solution its blue colour) are being continuously discharged at the cathode and removed from the solution, while they are not being replenished at the anode.
(d)(i) When active copper electrodes are used, the anode dissolves (decreases in mass/size) as copper atoms lose electrons to form copper(II) ions: \(Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-\). At the cathode, copper(II) ions are discharged, depositing a pink-brown layer of copper metal (increases in mass/size). (ii) The intensity of the blue colour remains constant because the concentration of \(Cu^{2+}\) ions in the solution does not change. The rate of copper dissolving at the anode is exactly equal to the rate of copper depositing at the cathode.
Marking scheme
(a) [2 marks total]: - (i) Copper [1] - (ii) Oxygen [1]
(b) [2 marks total]: - Correct species and balancing: \(4OH^- \rightarrow O_2 + 2H_2O + 4e^-\) - Correct state symbols: \((aq)\) for \(OH^-\), \((g)\) for \(O_2\), \((l)\) for \(H_2O\) [1]
(c) [2 marks total]: - Blue colour fades / solution becomes colourless [1] - Because \(Cu^{2+}\) ions are discharged / removed from the solution [1]
(d) [4 marks total]: - (i) Anode decreases in size/mass AND cathode increases in size/mass / gets coated in a pink-brown solid [1] - (ii) Concentration of \(Cu^{2+}\) ions remains constant [1] - (ii) Because rate of dissolving at the anode equals the rate of deposition at the cathode [1]
Question 4 · Structured Theory & Equation Completion
10 marks
A student investigates the rate of reaction between excess calcium carbonate chips and dilute hydrochloric acid.
(a) Draw a labelled diagram of the apparatus that could be used to measure the volume of gas produced over time. [3]
(b) State and explain, in terms of collision theory, how the rate of this reaction changes when:
(i) the temperature of the acid is increased [3]
(ii) the calcium carbonate chips are replaced with an equal mass of powdered calcium carbonate. [2]
(c) Describe the difference in both the initial rate of reaction (slope of the curve) and the final volume of gas produced if the experiment is repeated using hydrochloric acid of half the original concentration, keeping all other conditions constant. [2]
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Worked solution
(a) The diagram must depict a closed system where gas is generated and collected. Key components to label: conical flask containing reactant mixture, a tightly fitting rubber stopper, a delivery tube leading to a gas syringe or to an inverted measuring cylinder filled with water placed in a water trough.
(b)(i) Increasing the temperature increases the kinetic energy of the reactant particles. This causes them to move faster, increasing the frequency of collisions. More importantly, a much higher fraction of colliding particles possess energy equal to or greater than the activation energy (\(E_a\)), significantly increasing the frequency of successful collisions. (ii) Replacing chips with powder increases the total surface area of calcium carbonate exposed to the acid. This increases the frequency of collisions between reactant particles per unit time, thereby increasing the rate of reaction.
(c) Halving the concentration of the acid reduces the number of acid particles per unit volume, which decreases the frequency of collisions, resulting in a slower initial rate of reaction (the curve's gradient is less steep). Because calcium carbonate is in excess, hydrochloric acid is the limiting reactant. Halving its concentration halves the moles of acid reacted, which halves the final volume of carbon dioxide gas produced.
Marking scheme
(a) [3 marks total]: - Closed reaction vessel (conical flask with stopper) containing acid and solid [1] - Delivery tube connecting reaction vessel to collection apparatus [1] - Gas syringe / inverted graduated cylinder in water trough [1]
(b) [5 marks total]: - (i) Rate of reaction increases [1] - (i) More frequent collisions AND higher kinetic energy of particles [1] - (i) Greater proportion of particles have energy \(\ge E_a\) / more successful collisions per unit time [1] - (ii) Rate of reaction increases [1] - (ii) Greater surface area leads to more frequent collisions [1]
(c) [2 marks total]: - Initial rate is slower / shallower curve [1] - Final volume of gas is halved [1]
Question 5 · Structured Theory & Equation Completion
10 marks
The manufacture of sulfur trioxide in the Contact Process involves the reversible reaction:
\(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\) \(\Delta H = -197\) kJ/mol
(a) State the typical conditions of temperature, pressure, and catalyst used in this process. [3]
(b) Explain why a temperature of 450 °C is used rather than a much lower temperature, even though the forward reaction is exothermic. [3]
(c) State and explain the effect on the equilibrium yield of sulfur trioxide when:
(i) the pressure is increased [2]
(ii) a catalyst is added. [2]
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Worked solution
(a) The optimal conditions for the Contact Process are a temperature of 450 °C, a pressure of 1–2 atmospheres (or 100–200 kPa), and the presence of a vanadium(V) oxide (\(V_2O_5\)) catalyst.
(b) According to Le Chatelier's Principle, since the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the equilibrium position to the right, which would increase the equilibrium yield of \(SO_3\). However, at lower temperatures, the kinetic energy of the reactant molecules is very low, making the reaction rate too slow to be commercially viable. A temperature of 450 °C is used as a compromise between achieving a high rate of reaction and a high equilibrium yield.
(c)(i) Increasing the pressure causes the position of equilibrium to shift to the right-hand side. This is because there are fewer moles of gas on the right (2 moles of \(SO_3\)) than on the left (3 moles of reactants). The system opposes the increased pressure by shifting to the side with fewer gas molecules, thus increasing the yield of \(SO_3\). (ii) Adding a catalyst has no effect on the equilibrium yield. A catalyst lowers the activation energy of both the forward and backward reactions equally, thereby increasing their rates of reaction by the same factor without shifting the position of equilibrium.
Marking scheme
(a) [3 marks total]: - Temperature: 450 °C [1] - Pressure: 1–2 atmospheres (accept 1-2 bar / 100-200 kPa) [1] - Catalyst: vanadium(V) oxide / \(V_2O_5\) [1]
(b) [3 marks total]: - Lower temperature increases the yield because the forward reaction is exothermic [1] - Lower temperature makes the rate of reaction too slow [1] - 450 °C is a compromise temperature to give a reasonable rate and yield [1]
(c) [4 marks total]: - (i) Yield increases [1] - (i) Equilibrium shifts to the right because there are fewer gaseous moles on the right-hand side (2 vs 3) [1] - (ii) No change in yield [1] - (ii) Catalyst increases the rate of forward and reverse reactions equally [1]
Question 6 · Structured Theory & Equation Completion
10 marks
Magnesium sulfate is a soluble salt that can be prepared by reacting insoluble magnesium oxide with dilute sulfuric acid.
(a) Classify magnesium oxide as acidic, basic, amphoteric, or neutral. Give a reason for your answer. [2]
(b) Describe the experimental steps to prepare a pure, dry sample of magnesium sulfate crystals starting from magnesium oxide and dilute sulfuric acid. [5]
(c) Write the chemical equation for this reaction. [1]
(d) Name one other magnesium compound that could react with dilute sulfuric acid to prepare magnesium sulfate, and write the chemical equation for this reaction. [2]
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Worked solution
(a) Magnesium oxide is a metal oxide, which means it behaves as a base. It is classified as basic because it reacts with dilute sulfuric acid (an acid) to produce a salt (magnesium sulfate) and water, but does not react with bases.
(b) Preparation steps: 1. Measure a fixed volume of dilute sulfuric acid into a beaker and heat it gently. 2. Add magnesium oxide to the acid, stirring continuously, until no more dissolves (to ensure magnesium oxide is in excess and all acid is neutralised). 3. Filter the hot mixture using a funnel and filter paper to remove the unreacted, excess magnesium oxide (which remains as residue). 4. Transfer the filtrate (magnesium sulfate solution) to an evaporating basin and heat it until a saturated solution is formed (tested by seeing if crystals form on a cold glass rod). 5. Allow the hot saturated solution to cool slowly so that crystals form. Filter off the crystals, wash them with a small amount of cold distilled water, and dry them using filter paper.
(c) The balanced symbol equation is: \(MgO(s) + H_2SO_4(aq) \rightarrow MgSO_4(aq) + H_2O(l)\)
(d) Other suitable compounds: magnesium carbonate (\(MgCO_3\)), magnesium hydroxide (\(Mg(OH)_2\)), or magnesium metal (\(Mg\)). Equation for magnesium carbonate: \(MgCO_3 + H_2SO_4 \rightarrow MgSO_4 + H_2O + CO_2\)
Marking scheme
(a) [2 marks total]: - Basic [1] - It reacts with acids to form a salt and water / is a metal oxide [1]
(b) [5 marks total]: - Add excess magnesium oxide to dilute sulfuric acid [1] - Stir and warm/heat [1] - Filter the mixture to remove excess magnesium oxide [1] - Heat filtrate until saturated / to crystallisation point [1] - Leave to cool, filter off crystals, and dry them with filter paper / in a warm oven [1]
(d) [2 marks total]: - Name of compound: magnesium carbonate / magnesium hydroxide / magnesium metal [1] - Correct balanced equation [1]
Question 7 · Structured Theory & Equation Completion
10 marks
This question compares the bonding and properties of sodium chloride and chlorine gas.
(a) Sodium reacts with chlorine to form the ionic compound sodium chloride.
(i) Describe, in terms of electrons, how sodium atoms and chlorine atoms form sodium chloride. [2]
(ii) Draw a dot-and-cross diagram to show the electronic configuration and charges of the ions in sodium chloride. Show outer shell electrons only. [3]
(b) Chlorine is a diatomic gas, \(Cl_2\).
(i) Draw a dot-and-cross diagram to show the bonding in a molecule of chlorine. Show outer shell electrons only. [2]
(ii) Explain, in terms of structure and bonding, why chlorine has a very low boiling point (-34 °C) while sodium chloride has a very high melting point (801 °C). [3]
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Worked solution
(a)(i) During the reaction, a sodium atom (electronic configuration 2,8,1) transfers its single valence electron to a chlorine atom (electronic configuration 2,8,7). This electron transfer results in a sodium ion (\(Na^+\)) and a chloride ion (\(Cl^-\)), both having stable octet configurations. (ii) The sodium ion is represented by \([Na]^+\) (having lost its valence shell, no outer electrons need to be shown, or showing 8 electrons of the inner shell). The chloride ion is represented with square brackets, showing its 8 outer electrons (7 of its own, represented by crosses, and 1 gained from sodium, represented by a dot), with a minus charge outside the brackets: \([Cl]^-\).
(b)(i) Chlorine gas (\(Cl_2\)) is covalently bonded. Two chlorine atoms share one pair of electrons to achieve a stable noble gas configuration. In the overlap, show one dot and one cross. The remaining 6 valence electrons on each chlorine atom must be shown as non-bonding pairs. (ii) Chlorine consists of simple covalent molecules. While the covalent bonds within the molecules are strong, the forces between the molecules (intermolecular forces) are very weak and require very little thermal energy to overcome. Sodium chloride, on the other hand, exists as a giant ionic lattice. It contains strong electrostatic forces of attraction acting in all directions between oppositely charged ions (\(Na^+\) and \(Cl^-\)). Overcoming these strong forces requires a very large amount of thermal energy, resulting in a high melting point.
Marking scheme
(a) [5 marks total]: - (i) Sodium atom loses one electron AND chlorine atom gains one electron [1] - (i) To form \(Na^+\) and \(Cl^-\) ions [1] - (ii) \(Na^+\) shown with '+' charge and empty outer shell (or 8 electrons) [1] - (ii) \(Cl^-\) shown with 8 outer electrons (7 of one type, 1 of the other) and '-' charge [1] - (ii) Square brackets used correctly around both ions [1]
(b) [5 marks total]: - (i) One shared pair of electrons in the overlap area [1] - (i) 6 non-bonding electrons shown on each chlorine atom [1] - (ii) Chlorine has a simple molecular structure with weak intermolecular forces [1] - (ii) Sodium chloride has a giant ionic lattice with strong electrostatic forces between oppositely charged ions [1] - (ii) Breaking the strong ionic bonds requires significantly more energy than overcoming weak intermolecular forces [1]
Question 8 · structured
10 marks
A student heats a sample of hydrated cobalt(II) nitrate, \(\text{Co(NO}_3)_2 \cdot x\text{H}_2\text{O}\), to determine the value of \(x\).
The student records the following results: - mass of empty crucible = \(20.00\text{ g}\) - mass of crucible + hydrated cobalt(II) nitrate = \(25.82\text{ g}\) - mass of crucible + anhydrous cobalt(II) nitrate (after gentle heating to constant mass) = \(23.66\text{ g}\)
(a) Use the results to calculate the value of \(x\) in \(\text{Co(NO}_3)_2 \cdot x\text{H}_2\text{O}\). Show your working. [4]
(b) When anhydrous cobalt(II) nitrate is heated strongly, it decomposes according to the following equation. Balance the equation: $$\dots\text{Co(NO}_3)_2(s) \rightarrow \dots\text{CoO}(s) + \dots\text{NO}_2(g) + \text{O}_2(g)$$ [1]
(c) Calculate the total volume of gas, in \(\text{dm}^3\) at r.t.p., produced when the \(3.66\text{ g}\) of anhydrous cobalt(II) nitrate from the student's experiment is completely decomposed. [3]
(d) The oxygen gas produced in (c) is collected and reacted with \(0.60\text{ dm}^3\) of carbon monoxide gas, \(\text{CO}\), to form carbon dioxide, \(\text{CO}_2\). $$2\text{CO}(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g)$$ Identify the limiting reactant and calculate the volume of carbon dioxide, in \(\text{dm}^3\), produced at r.t.p. [2]
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Worked solution
(a) Mass of hydrated salt = \(25.82 - 20.00 = 5.82\text{ g}\) Mass of anhydrous salt = \(23.66 - 20.00 = 3.66\text{ g}\) Mass of water lost = \(5.82 - 3.66 = 2.16\text{ g}\) \(M_r\) of \(\text{Co(NO}_3)_2 = 59 + 2(14 + (16 \times 3)) = 183\) \(\text{Moles of anhydrous salt} = \frac{3.66}{183} = 0.02\text{ mol}\) \(\text{Moles of water} = \frac{2.16}{18} = 0.12\text{ mol}\) \(\text{Ratio of Co(NO}_3)_2 : \text{H}_2\text{O} = 0.02 : 0.12 = 1 : 6\) Therefore, \(x = 6\).
(c) \(\text{Moles of anhydrous salt decomposed} = 0.02\text{ mol}\) From the balanced equation, \(2\text{ mol}\) of \(\text{Co(NO}_3)_2\) produces \(5\text{ mol}\) of total gas (\(4\text{ mol } \text{NO}_2 + 1\text{ mol } \text{O}_2\)). \(\text{Total moles of gas produced} = 0.02 \times \frac{5}{2} = 0.05\text{ mol}\) \(\text{Total volume of gas} = 0.05 \times 24 = 1.2\text{ dm}^3\).
(d) \(\text{Moles of O}_2\) produced in (c) = \(0.01\text{ mol}\) \(\text{Volume of O}_2 = 0.01 \times 24 = 0.24\text{ dm}^3\) Reacting ratio: \(2\text{CO} : 1\text{O}_2\) \(0.24\text{ dm}^3\) of \(\text{O}_2\) requires \(0.48\text{ dm}^3\) of \(\text{CO}\). Since \(0.60\text{ dm}^3\) of \(\text{CO}\) is present, \(\text{CO}\) is in excess and \(\text{O}_2\) is the limiting reactant. \(\text{Volume of CO}_2\) produced = \(2 \times 0.24\text{ dm}^3 = 0.48\text{ dm}^3\).
Marking scheme
(a) [4 marks total] - M1: Calculate mass of water lost (\(2.16\text{ g}\)) AND mass of anhydrous salt (\(3.66\text{ g}\)) [1] - M2: Calculate moles of anhydrous \(\text{Co(NO}_3)_2 = 0.02\text{ mol}\) (using \(M_r = 183\)) [1] - M3: Calculate moles of \(\text{H}_2\text{O} = 0.12\text{ mol}\) [1] - M4: Deduce ratio of \(1 : 6\) to find \(x = 6\) [1]
(c) [3 marks total] - M1: State/use moles of anhydrous \(\text{Co(NO}_3)_2 = 0.02\text{ mol}\) (allow ecf) [1] - M2: Use molar ratio from (b) to find total moles of gas = \(0.05\text{ mol}\) (or moles of \(\text{NO}_2 = 0.04\) and moles of \(\text{O}_2 = 0.01\)) [1] - M3: Calculate total volume = \(1.2\text{ dm}^3\) (or \(1200\text{ cm}^3\)) [1]
(d) [2 marks total] - M1: Show by calculation that oxygen is the limiting reactant (e.g. \(0.24\text{ dm}^3\) of \(\text{O}_2\) requires only \(0.48\text{ dm}^3\) of \(\text{CO}\)) [1] - M2: Calculate volume of \(\text{CO}_2\) produced = \(0.48\text{ dm}^3\) (allow ecf) [1]
Paper 6 Alternative to Practical
Answer all questions. Show measurements to the appropriate decimal resolution and plan experimental procedures.
5 Question · 46 marks
Question 1 · theory
12 marks
A student investigates the reaction between dilute hydrochloric acid and aqueous potassium hydroxide.
In each of six experiments, the student: - measures 25.0 cm³ of dilute hydrochloric acid into a polystyrene cup - records the initial temperature of the acid - adds a specific volume of aqueous potassium hydroxide to the cup, stirs the mixture and records the maximum temperature reached.
(a) Complete Table 1.1 by calculating the missing values: [i] temperature rise for Experiment 1 [ii] maximum temperature for Experiment 3 [iii] temperature rise for Experiment 5 [3]
(b) Describe how the temperature rise changes as the volume of potassium hydroxide increases from 5.0 cm³ to 30.0 cm³. [2]
(c) Explain why the temperature decreases after the maximum temperature rise is reached (when more than 20.0 cm³ of potassium hydroxide is added). [2]
(d) Suggest one advantage of using a polystyrene cup instead of a glass beaker in this investigation. [1]
(e) Suggest one change to the apparatus that would improve the accuracy of the temperature measurements. [1]
(f) Describe a chemical test to show that the solution in Experiment 6 is alkaline. State the observation for a positive result. [3]
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Worked solution
Detailed step-by-step breakdown:
(a) Calculations: - [i] Temperature rise = Maximum temperature - Initial temperature = 22.3 - 19.5 = 2.8 °C. - [ii] Maximum temperature = Initial temperature + Temperature rise = 19.5 + 8.4 = 27.9 °C. - [iii] Temperature rise = Maximum temperature - Initial temperature = 28.5 - 19.5 = 9.0 °C.
(b) Trend analysis: - As volume increases from 5.0 to 20.0 cm³, the temperature rise increases to a maximum of 11.0 °C. - As volume increases further from 20.0 to 30.0 cm³, the temperature rise decreases.
(c) Explanation: - The reaction is fully complete (all acid has been neutralised) when 20.0 cm³ of KOH has been added. No more heat is generated after this point. - Adding extra cold KOH solution simply dilutes and cools down the warm reaction mixture.
(d) Material choice: - Polystyrene is a heat insulator, which reduces heat transfer/loss to the surroundings, making the measured maximum temperatures more accurate.
(e) Apparatus improvement: - Use a digital thermometer or a thermometer with smaller scale divisions (e.g., to 0.1 °C) to measure temperatures more precisely. Alternatively, use a plastic lid to reduce heat loss from the top.
(f) Chemical test: - Test: Add a few drops of phenolphthalein indicator or use red litmus paper. - Observation: Phenolphthalein turns pink (or red litmus turns blue).
Marking scheme
(a) [3 marks] - 1 mark for [i] = 2.8 - 1 mark for [ii] = 27.9 (accept 27.9 or 28) - 1 mark for [iii] = 9.0 (accept 9)
(b) [2 marks] - 1 mark for stating that it increases to a maximum (at 20.0 cm³) - 1 mark for stating that it decreases / falls after 20.0 cm³
(c) [2 marks] - 1 mark for stating the acid is fully neutralised / reaction is complete - 1 mark for stating that cold KOH / solution is added which cools the mixture down
(d) [1 mark] - 1 mark for stating polystyrene is a better heat insulator / reduces heat loss to the surroundings (Reject: 'prevents heat loss')
(e) [1 mark] - 1 mark for any correct improvement, e.g., use a thermometer with a higher resolution / digital thermometer / use a lid on the cup to prevent heat loss / insulate the cup further.
(f) [3 marks] - 1 mark for named indicator / test paper, e.g., red litmus paper / phenolphthalein / universal indicator - 1 mark for describing the test procedure, e.g., dip the paper into the solution / add a few drops of the indicator - 1 mark for the correct colour change, e.g., turns blue (for red litmus) / turns pink (for phenolphthalein) / turns blue-purple (for universal indicator)
Question 2 · theory
12 marks
A student investigates the rate of reaction between dilute nitric acid and excess calcium carbonate chips at room temperature.
The reaction produces carbon dioxide gas, which is collected in a gas syringe. The volume of carbon dioxide gas collected is recorded every 30 seconds.
(a) Complete Table 2.1 by calculating the missing values: [i] increase in volume of CO₂ in the 0-30 s interval [ii] total volume of CO₂ at 90 s [iii] increase in volume of CO₂ in the 120-150 s interval [3]
(b) State how the rate of the reaction changes during the 150 seconds. Explain your answer with reference to the data in Table 2.1. [2]
(c) Explain why the reaction rate decreases over time. [1]
(d) The student repeats the experiment using the same mass of calcium carbonate but in the form of a fine powder instead of chips. All other conditions are kept constant. (i) State how the rate of reaction would change. [1] (ii) Describe how a graph of volume of gas against time for the powder would compare to the curve for the chips. [2]
(e) Describe a chemical test to confirm that the gas produced is carbon dioxide. State the observation for a positive result. [3]
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Worked solution
Detailed step-by-step breakdown:
(a) Calculations: - [i] Increase in volume = Volume at 30 s - Volume at 0 s = 16 - 0 = 16 cm³. - [ii] Total volume at 90 s = Volume at 60 s + Increase = 28 + 8 = 36 cm³. - [iii] Increase in volume = Volume at 150 s - Volume at 120 s = 42 - 40 = 2 cm³.
(b) Trend analysis: - The rate of the reaction decreases continuously. This is shown by the decreasing volume increase in each 30-second interval: 16 cm³ in the first 30 seconds, then dropping to 12 cm³, 8 cm³, 4 cm³, and finally 2 cm³.
(c) Explanation: - As the reaction proceeds, the nitric acid is consumed, meaning its concentration decreases. This leads to fewer successful collisions per unit time between the reactant particles, slowing down the reaction.
(d) Comparing powder and chips: - (i) Using powder increases the surface area of the calcium carbonate, which increases the rate of reaction. - (ii) On a graph of volume vs. time, the curve for the powder would have a steeper initial gradient (slope) because the reaction is faster. It would level off sooner than the chips curve but at exactly the same final volume (42 cm³) because the mass of calcium carbonate (limiting/excess considerations) and nitric acid volume remain unchanged.
(e) Carbon dioxide gas test: - Test: Bubble the gas into limewater (calcium hydroxide solution). - Observation: The limewater turns milky/cloudy/precipitate forms.
Marking scheme
(a) [3 marks] - 1 mark for [i] = 16 - 1 mark for [ii] = 36 - 1 mark for [iii] = 2
(b) [2 marks] - 1 mark for stating that the rate decreases over time - 1 mark for referencing the decreasing intervals in the table (e.g., volume gain drops from 16 to 2 cm³)
(c) [1 mark] - 1 mark for stating that the acid / reactants are used up / concentration decreases (Accept: fewer reactant particles left to collide)
(d) [3 marks] - (i) 1 mark for stating the rate of reaction increases / faster reaction - (ii) 2 marks: 1 mark for stating the curve has a steeper gradient / slope / levels off earlier; 1 mark for stating it levels off at the same final volume (42 cm³)
(e) [3 marks] - 1 mark for bubble the gas / pass the gas through - 1 mark for limewater / aqueous calcium hydroxide - 1 mark for correct observation: turns cloudy / milky / white precipitate forms
Question 3 · structured
10 marks
A student carries out tests on solid **Y**, which is iron(II) sulfite.
**(a)** State the colour of solid **Y**. [1]
**(b)** The student dissolves solid **Y** in distilled water to form solution **Y**.
**(i)** To the first portion of solution **Y**, the student adds aqueous sodium hydroxide dropwise, and then in excess. State the expected observations. * adding dropwise: ......................... * in excess: ......................... [2]
**(ii)** To the second portion of solution **Y**, the student adds aqueous ammonia dropwise, and then in excess. State the expected observations. * adding dropwise: ......................... * in excess: ......................... [2]
**(c)** To a third portion of solution **Y**, the student adds dilute hydrochloric acid and warms the mixture. A gas is evolved.
**(i)** State the chemical test used to identify this gas. [2]
**(ii)** State the observation that confirms the presence of this gas. [1]
**(d)** Identify the two ions present in solid **Y**. * cation: ......................... * anion: ......................... [2]
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Worked solution
**(a)** Transition metal compounds of iron(II) are typically green. Therefore, the colour of solid **Y** is green or pale green.
**(b)(i)** According to the qualitative analysis notes, adding aqueous sodium hydroxide to a solution containing iron(II) ions (\( \text{Fe}^{2+} \)) produces a green precipitate that is insoluble in excess.
**(b)(ii)** Adding aqueous ammonia to a solution containing iron(II) ions produces a green precipitate that is also insoluble in excess.
**(c)(i)** Sulfite ions (\( \text{SO}_3^{2-} \)) react with dilute acid to produce sulfur dioxide gas (\( \text{SO}_2 \)). The standard test for sulfur dioxide gas is passing it through or exposing it to acidified potassium manganate(VII).
**(c)(ii)** Sulfur dioxide is a reducing agent and reduces the purple manganate(VII) ions to colourless manganese(II) ions. Thus, the purple colour turns colourless.
**(d)** The cation is iron(II) (\( \text{Fe}^{2+} \)) and the anion is sulfite (\( \text{SO}_3^{2-} \)).
Marking scheme
**(a)** Green / pale green [1] (Reject: blue-green / yellow-green)
**(b)(i)** dropwise: green precipitate [1] in excess: insoluble / precipitate remains [1] (Ignore: turns brown near surface on standing)
**(b)(ii)** dropwise: green precipitate [1] in excess: insoluble / precipitate remains [1]
**(c)(i)** (acidified) [1] potassium manganate(VII) (solution or paper) [1]
**(c)(ii)** turns from purple to colourless / decolourises [1] (Reject: turns clear / transparent)
Three different brands of solid antacid tablets, Brand X, Brand Y, and Brand Z, are used to neutralise excess stomach acid. You are provided with these tablets, dilute hydrochloric acid of a known concentration, methyl orange indicator, and common laboratory apparatus. Plan an investigation to determine which brand of antacid tablet is the most effective at neutralising the acid per gram of tablet. In your plan, you should describe: how you will perform the experiment, the measurements you will record, how you will make it a fair test, and how you will use your results to find the most effective antacid.
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Worked solution
First, crush the antacid tablets of Brand X, Brand Y, and Brand Z separately using a pestle and mortar. Use a digital balance to weigh out equal masses, for example 1.0 g, of each crushed tablet. Transfer the 1.0 g of Brand X powder into a clean conical flask and add 25 cm3 of distilled water using a measuring cylinder to suspend the solid. Add 3 to 5 drops of methyl orange indicator to the flask, which will turn yellow. Fill a burette with the provided dilute hydrochloric acid and record the initial burette reading. Slowly run the acid from the burette into the conical flask while continuously swirling the flask. Stop adding the acid as soon as the indicator changes color from yellow to orange/red, and record the final burette reading. Calculate the volume of acid used. Repeat this entire procedure using 1.0 g of Brand Y and 1.0 g of Brand Z. The brand of antacid tablet that requires the largest volume of hydrochloric acid to reach the end-point is the most effective at neutralising the acid per gram.
Marking scheme
Award up to 6 marks for the following points: 1. Crush the tablets and weigh out an equal mass of each brand of tablet using a balance. 2. Place the weighed sample into a conical flask and add a measured volume of distilled water using a measuring cylinder. 3. Add a few drops of methyl orange indicator to the mixture. 4. Run the dilute hydrochloric acid from a burette into the flask while swirling until the color changes from yellow to orange/red. 5. Repeat the titration procedure for the other two brands of antacid tablets using the same mass of tablet. 6. Conclude that the brand requiring the largest volume of hydrochloric acid is the most effective per gram.
Question 5 · practical
6 marks
Three different brands of solid antacid tablets, Brand X, Brand Y, and Brand Z, are used to neutralise excess stomach acid. You are provided with these tablets, dilute hydrochloric acid of a known concentration, methyl orange indicator, and common laboratory apparatus. Plan an investigation to determine which brand of antacid tablet is the most effective at neutralising the acid per gram of tablet. In your plan, you should describe: how you will perform the experiment, the measurements you will record, how you will make it a fair test, and how you will use your results to find the most effective antacid.
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Worked solution
First, crush the antacid tablets of Brand X, Brand Y, and Brand Z separately using a pestle and mortar. Use a digital balance to weigh out equal masses, for example 1.0 g, of each crushed tablet. Transfer the 1.0 g of Brand X powder into a clean conical flask and add 25 cm3 of distilled water using a measuring cylinder to suspend the solid. Add 3 to 5 drops of methyl orange indicator to the flask, which will turn yellow. Fill a burette with the provided dilute hydrochloric acid and record the initial burette reading. Slowly run the acid from the burette into the conical flask while continuously swirling the flask. Stop adding the acid as soon as the indicator changes color from yellow to orange/red, and record the final burette reading. Calculate the volume of acid used. Repeat this entire procedure using 1.0 g of Brand Y and 1.0 g of Brand Z. The brand of antacid tablet that requires the largest volume of hydrochloric acid to reach the end-point is the most effective at neutralising the acid per gram.
Marking scheme
Award up to 6 marks for the following points: 1. Crush the tablets and weigh out an equal mass of each brand of tablet using a balance. 2. Place the weighed sample into a conical flask and add a measured volume of distilled water using a measuring cylinder. 3. Add a few drops of methyl orange indicator to the mixture. 4. Run the dilute hydrochloric acid from a burette into the flask while swirling until the color changes from yellow to orange/red. 5. Repeat the titration procedure for the other two brands of antacid tablets using the same mass of tablet. 6. Conclude that the brand requiring the largest volume of hydrochloric acid is the most effective per gram.
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