An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V1) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.
Paper 11 (Multiple Choice)
There are forty questions on this paper. Answer all questions. For each question there are four possible answers.
40 PastPaper.question · 40 PastPaper.marks
PastPaper.question 1 · multiple-choice
1 PastPaper.marks
At room temperature and pressure (r.t.p.), \(10\text{ cm}^3\) of a gaseous hydrocarbon \(C_xH_y\) was exploded with \(100\text{ cm}^3\) of oxygen (an excess). After cooling to r.t.p., the total volume of gas remaining was \(85\text{ cm}^3\). On shaking this gas mixture with aqueous sodium hydroxide, the volume of gas decreased to \(45\text{ cm}^3\).
What is the molecular formula of the hydrocarbon?
A.C4H10
B.C4H8
C.C4H6
D.C3H8
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PastPaper.workedSolution
1. Shaking with aqueous NaOH absorbs \(CO_2\). The decrease in volume is \(85\text{ cm}^3 - 45\text{ cm}^3 = 40\text{ cm}^3\), which is the volume of \(CO_2\) produced. 2. Since \(10\text{ cm}^3\) of \(C_xH_y\) produces \(40\text{ cm}^3\) of \(CO_2\), the value of \(x\) is \(\frac{40}{10} = 4\). 3. The remaining gas after NaOH treatment (\(45\text{ cm}^3\)) is the unreacted (excess) \(O_2\). 4. The volume of \(O_2\) reacted is \(100\text{ cm}^3 - 45\text{ cm}^3 = 55\text{ cm}^3\). 5. From the combustion equation: \(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O(l)\). 6. The ratio of \(O_2\) reacted to hydrocarbon is \(\frac{55}{10} = 5.5\). Therefore, \(x + \frac{y}{4} = 5.5\). 7. Substituting \(x = 4\) into the equation: \(4 + \frac{y}{4} = 5.5 \Rightarrow \frac{y}{4} = 1.5 \Rightarrow y = 6\). 8. Thus, the molecular formula is \(C_4H_6\).
PastPaper.markingScheme
Award 1 mark for the correct answer C. - 1 mark: Correctly identifies \(C_4H_6\) through calculation of \(CO_2\) volume (\(40\text{ cm}^3\)) and \(O_2\) consumed (\(55\text{ cm}^3\)) to find \(x = 4\) and \(y = 6\).
PastPaper.question 2 · multiple-choice
1 PastPaper.marks
A \(1.00\text{ g}\) sample of a Group 2 metal \(M\) reacts completely with excess dilute hydrochloric acid.
The reaction produces \(600\text{ cm}^3\) of hydrogen gas, measured at room temperature and pressure (r.t.p.).
What is the identity of metal \(M\)? [Molar volume of gas at r.t.p. is \(24.0\text{ dm}^3\text{ mol}^{-1}\)]
A.magnesium
B.calcium
C.strontium
D.barium
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PastPaper.workedSolution
1. Find the moles of \(H_2\) gas produced: \(\text{moles of } H_2 = \frac{600\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.025\text{ mol}\). 2. From the stoichiometry of the equation, 1 mole of \(M\) produces 1 mole of \(H_2\). Thus, \(\text{moles of } M = 0.025\text{ mol}\). 3. Calculate the relative atomic mass (\(A_r\)) of \(M\): \(A_r(M) = \frac{\text{mass}}{\text{moles}} = \frac{1.00\text{ g}}{0.025\text{ mol}} = 40.0\text{ g mol}^{-1}\). 4. Comparing with the Periodic Table, the Group 2 metal with \(A_r \approx 40.1\text{ g mol}^{-1}\) is calcium.
PastPaper.markingScheme
Award 1 mark for the correct answer B. - 1 mark: Correctly calculates the moles of \(H_2\) (\(0.025\text{ mol}\)), links it to the moles of \(M\), and computes \(A_r = 40.0\text{ g mol}^{-1}\) to identify calcium.
PastPaper.question 3 · multiple-choice
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An oxide of a Period 3 element, \(X\), is a white solid with a high melting point. It is insoluble in water and does not react with dilute hydrochloric acid, but it reacts with hot, concentrated aqueous sodium hydroxide.
What is the element \(X\)?
A.aluminium
B.silicon
C.phosphorus
D.sulfur
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PastPaper.workedSolution
1. The oxide of silicon is silicon dioxide, \(SiO_2\), which is a giant covalent macromolecule with a very high melting point. 2. Being acidic but macromolecular, it does not dissolve in or react with water, nor does it react with acids like \(HCl(aq)\). 3. It does, however, react with hot, concentrated alkali such as \(NaOH(aq)\) to form soluble sodium silicate: \(SiO_2(s) + 2NaOH(aq) \rightarrow Na_2SiO_3(aq) + H_2O(l)\). 4. Aluminium oxide, \(Al_2O_3\), is amphoteric and would react with both dilute \(HCl\) and \(NaOH\). 5. Phosphorus and sulfur oxides are simple molecular structures with low melting points.
PastPaper.markingScheme
Award 1 mark for the correct answer B. - 1 mark: Correctly deduces that silicon dioxide is the giant covalent oxide that behaves as a weak acid, reacting only with concentrated alkali and not with acids.
PastPaper.question 4 · multiple-choice
1 PastPaper.marks
Which sequence of Period 3 ions is written in order of decreasing ionic radius?
A.P3-, S2-, Cl-, Na+
B.Na+, Mg2+, Al3+, P3-
C.Cl-, S2-, P3-, Si4+
D.Si4+, Al3+, Mg2+, Na+
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PastPaper.workedSolution
1. For the anions \(P^{3-}\), \(S^{2-}\), and \(Cl^-\), they are isoelectronic, each having 18 electrons (electronic configuration \(2, 8, 8\)). As the nuclear charge increases from +15 (P) to +17 (Cl), the outer electrons are attracted more strongly, resulting in a decrease in ionic radius: \(P^{3-} > S^{2-} > Cl^-\). 2. \(Na^+\) has only 10 electrons (electronic configuration \(2, 8\)) distributed over two shells, so its ionic radius is significantly smaller than the anions which have three occupied shells. 3. Therefore, the sequence \(P^{3-} > S^{2-} > Cl^- > Na^+\) shows a continuous decrease in ionic radius.
PastPaper.markingScheme
Award 1 mark for the correct answer A. - 1 mark: Identifies that the ionic radius decreases across isoelectronic anions as nuclear charge increases, and that the cation with fewer shells is smaller than the anions.
PastPaper.question 5 · multiple-choice
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An aqueous solution of cobalt(II) sulfate is pink due to the presence of \([Co(H_2O)_6]^{2+}\) ions. When excess concentrated hydrochloric acid is added, the solution turns deep blue.
Which statement correctly explains this observation?
A.The coordination number of the cobalt ion increases from 4 to 6.
B.The oxidation state of the cobalt ion changes from +2 to +4.
C.The geometry around the cobalt ion changes from octahedral to tetrahedral.
D.Water molecules act as stronger ligands than chloride ions, displacing them.
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PastPaper.workedSolution
1. The pink starting complex is \([Co(H_2O)_6]^{2+}\), which is octahedral with a coordination number of 6. 2. Adding excess concentrated \(HCl\) introduces a high concentration of \(Cl^-\) ligands, leading to ligand exchange to form the blue tetrachlorocobaltate(II) complex, \([CoCl_4]^{2-}\). 3. In \([CoCl_4]^{2-}\), the coordination number is 4 and the geometry is tetrahedral. 4. Therefore, the coordination number decreases from 6 to 4, and the geometry changes from octahedral to tetrahedral. The oxidation state of cobalt remains +2 in both species.
PastPaper.markingScheme
Award 1 mark for the correct answer C. - 1 mark: Correctly identifies the transition from the octahedral hexaaqua complex to the tetrahedral tetrachloro complex during ligand exchange with concentrated hydrochloric acid.
PastPaper.question 6 · multiple-choice
1 PastPaper.marks
Which of the following transition metal complexes can exist as a pair of optical isomers (enantiomers)?
[where 'en' represents the bidentate ligand ethane-1,2-diamine]
A.cis-[Co(en)2Cl2]+
B.trans-[Co(en)2Cl2]+
C.[Co(NH3)4Cl2]+
D.[Co(en)Cl4]-
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PastPaper.workedSolution
1. Optical isomerism occurs in octahedral complexes containing bidentate ligands when the isomer lacks a plane of symmetry. 2. \(\text{trans-}[Co(en)_2Cl_2]^+\) has a plane of symmetry passing through the metal center and the chloride ligands, making it achiral. 3. \(\text{cis-}[Co(en)_2Cl_2]^+\) lacks any plane of symmetry and therefore has a non-superimposable mirror image, allowing it to exist as a pair of optical isomers. 4. Both \([Co(NH_3)_4Cl_2]^+\) and \([Co(en)Cl_4]^-\) have planes of symmetry and are achiral.
PastPaper.markingScheme
Award 1 mark for the correct answer A. - 1 mark: Correctly identifies that the cis-isomer of the bis-ethylenediamine cobalt complex lacks a plane of symmetry and is therefore chiral (optical isomers exist).
PastPaper.question 7 · multiple-choice
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The standard enthalpy changes of combustion, \(\Delta H^\ominus_c\), for carbon, hydrogen, and propanoic acid are given below:
What is the standard enthalpy change of formation, \(\Delta H^\ominus_f\), of propanoic acid?
A.-513 kJ mol-1
B.+513 kJ mol-1
C.-1147 kJ mol-1
D.-2040 kJ mol-1
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PastPaper.workedSolution
1. Write the chemical equation for the standard enthalpy of formation of propanoic acid: \(3C(graphite) + 3H_2(g) + O_2(g) \rightarrow C_2H_5CO_2H(l)\) 2. Apply Hess's law using enthalpy changes of combustion: \(\Delta H^\ominus_f = \sum \Delta H^\ominus_c(\text{reactants}) - \sum \Delta H^\ominus_c(\text{products})\) 3. Calculate the sum of combustion enthalpies for the reactants: \(\sum \Delta H^\ominus_c(\text{reactants}) = 3 \times \Delta H^\ominus_c[C(graphite)] + 3 \times \Delta H^\ominus_c[H_2(g)]\) \(= 3(-394) + 3(-286) = -1182 - 858 = -2040\text{ kJ mol}^{-1}\). 4. Subtract the combustion enthalpy of the product: \(\Delta H^\ominus_f = -2040 - (-1527) = -513\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
Award 1 mark for the correct answer A. - 1 mark: Performs correct stoichiometry multiplication (x3 for carbon and hydrogen) and correctly applies Hess's law subtraction to find \(-513\text{ kJ mol}^{-1}\).
PastPaper.question 8 · multiple-choice
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Nitrogen dioxide, \(NO_2\), is a brown gas that exists in dynamic equilibrium with colorless dinitrogen tetroxide, \(N_2O_4\).
\(2NO_2(g) \rightleftharpoons N_2O_4(g) \quad \Delta H = -57\text{ kJ mol}^{-1}\)
A mixture of these two gases is at equilibrium in a sealed gas syringe at room temperature. The plunger of the syringe is suddenly pushed in, reducing the volume of the gas mixture, and the plunger is held in this position.
What describes the immediate color change and the subsequent color change of the gas mixture?
A.Immediate: becomes darker brown; Subsequent: becomes lighter brown
B.Immediate: becomes darker brown; Subsequent: becomes even darker brown
C.Immediate: becomes lighter brown; Subsequent: becomes darker brown
D.Immediate: remains unchanged; Subsequent: becomes lighter brown
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PastPaper.workedSolution
1. When the volume is suddenly decreased, the concentrations of all gas molecules (\(NO_2\) and \(N_2O_4\)) increase immediately because they are compressed into a smaller space. 2. Since \(NO_2\) is brown, this sudden increase in concentration immediately intensifies the brown color (it becomes darker brown). 3. Subsequently, according to Le Chatelier's principle, the system responds to the increased pressure/concentration by shifting the position of equilibrium to the side with fewer gas molecules (the right-hand side, to form more colorless \(N_2O_4\)). 4. As \(NO_2\) is consumed to form \(N_2O_4\), the concentration of \(NO_2\) decreases from its peak, causing the brown color to subsequently fade (it becomes lighter brown).
PastPaper.markingScheme
Award 1 mark for the correct answer A. - 1 mark: Correctly identifies that compression immediately increases concentration (causing the mixture to darken) and subsequent equilibrium shift to the right decreases \(NO_2\) concentration (causing the mixture to lighten).
PastPaper.question 9 · multiple-choice
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How many total ions are present in \(25.0\text{ cm}^3\) of a \(0.150\text{ mol dm}^{-3}\) solution of aluminium sulfate, \(\text{Al}_2(\text{SO}_4)_3\)?
A.\(2.26 \times 10^{21}\)
B.\(4.52 \times 10^{21}\)
C.\(6.77 \times 10^{21}\)
D.\(1.13 \times 10^{22}\)
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PastPaper.workedSolution
Moles of \(\text{Al}_2(\text{SO}_4)_3 = 0.0250\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 3.75 \times 10^{-3}\text{ mol}\). One formula unit of \(\text{Al}_2(\text{SO}_4)_3\) dissociates to form five ions: two \(\text{Al}^{3+}\) ions and three \(\text{SO}_4^{2-}\) ions. Thus, total moles of ions = \(5 \times 3.75 \times 10^{-3} = 0.01875\text{ mol}\). Total number of ions = \(0.01875 \times 6.02 \times 10^{23} = 1.13 \times 10^{22}\).
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1 mark for the correct option D.
PastPaper.question 10 · multiple-choice
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A solid mixture of anhydrous sodium carbonate (\(\text{Na}_2\text{CO}_3\), \(M_r = 106.0\)) and sodium hydrogencarbonate (\(\text{NaHCO}_3\), \(M_r = 84.0\)) has a mass of \(5.00\text{ g}\). When heated strongly, the sodium hydrogencarbonate completely decomposes to form sodium carbonate, water, and carbon dioxide: \(2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{g}) + \text{CO}_2(\text{g})\). The sodium carbonate does not decompose. After heating, the remaining dry solid has a mass of \(4.07\text{ g}\). What is the percentage by mass of sodium hydrogencarbonate in the original mixture?
A.18.6%
B.31.0%
C.50.4%
D.74.8%
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PastPaper.workedSolution
Let \(x\) be the mass of \(\text{NaHCO}_3\) in the mixture. Mass loss = \(5.00 - 4.07 = 0.93\text{ g}\). The decomposition of \(2\text{ mol}\) of \(\text{NaHCO}_3\) (\(168.0\text{ g}\)) produces a mass loss of \(62.0\text{ g}\) (due to water and carbon dioxide gases). Therefore, the mass of \(\text{NaHCO}_3\) is \(0.93 \times (168.0 / 62.0) = 2.52\text{ g}\). The percentage by mass of \(\text{NaHCO}_3\) is \((2.52 / 5.00) \times 100\% = 50.4\%\).
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1 mark for the correct option C.
PastPaper.question 11 · multiple-choice
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Which statement about the reactions of Period 3 elements with water is correct?
A.Silicon reacts with cold water to produce a solution of silicon dioxide.
B.Magnesium reacts extremely slowly with cold water to form an alkaline solution and a gas.
C.Phosphorus reacts with steam to produce phosphoric(V) acid and hydrogen.
D.Chlorine reacts with water in a redox reaction to form a solution with a pH of 7.
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PastPaper.workedSolution
Magnesium reacts extremely slowly with cold water to form sparingly soluble magnesium hydroxide (which makes the solution weakly alkaline) and hydrogen gas. Silicon does not react with cold water; phosphorus does not react with steam; chlorine reacts with water to form a strongly acidic solution of \(\text{HCl}\) and \(\text{HClO}\) with a pH of around 1-2.
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1 mark for the correct option B.
PastPaper.question 12 · multiple-choice
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Two Period 3 elements, \(\text{A}\) and \(\text{B}\), react with chlorine to form chlorides \(\text{X}\) and \(\text{Y}\) respectively. Chloride \(\text{X}\) dissolves in water to form a solution with a pH of approximately 7. Chloride \(\text{Y}\) reacts violently with water to form a strongly acidic solution with pH less than 2, while releasing white fumes. Which pair could represent elements \(\text{A}\) and \(\text{B}\)?
A.A is magnesium; B is aluminium
B.A is sodium; B is silicon
C.A is phosphorus; B is sulfur
D.A is aluminium; B is sodium
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PastPaper.workedSolution
Chloride \(\text{X}\) is \(\text{NaCl}\) (formed from sodium), which dissolves in water to form a neutral solution (pH ~ 7). Chloride \(\text{Y}\) is \(\text{SiCl}_4\) (formed from silicon), which reacts violently with water to form \(\text{SiO}_2\) and \(\text{HCl}\) (strongly acidic, pH < 2, releasing white fumes of \(\text{HCl}\)).
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1 mark for the correct option B.
PastPaper.question 13 · multiple-choice
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The addition of concentrated hydrochloric acid to an aqueous solution containing \([\text{Co(H}_2\text{O)}_6]^{2+}\) ions results in a color change from pink to blue. Which statement correctly explains this observation?
A.The cobalt(II) ion is reduced to cobalt(I).
B.The coordination number of the cobalt ion increases from 4 to 6.
C.A ligand exchange reaction occurs, resulting in a change in the shape of the complex from octahedral to tetrahedral.
D.The chloride ligands are stronger field ligands than water, causing a larger d-orbital splitting.
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PastPaper.workedSolution
The addition of concentrated hydrochloric acid results in the formation of the tetrahedral complex \([\text{CoCl}_4]^{2-}\) (blue) from the octahedral complex \([\text{Co(H}_2\text{O)}_6]^{2+}\) (pink). This is a ligand exchange reaction where the coordination number decreases from 6 to 4, and the shape changes from octahedral to tetrahedral.
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1 mark for the correct option C.
PastPaper.question 14 · multiple-choice
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Which statement correctly explains why many transition metal complex ions are coloured?
A.Light is emitted when d-electrons drop from d-orbitals of higher energy to d-orbitals of lower energy.
B.The d-orbitals are split into two groups of different energies by the ligands; d-d electron transition absorbs a specific frequency of visible light, and the complementary colour is transmitted.
C.The d-orbitals are split into two groups of different energies by the ligands; d-d electron transition emits a specific frequency of visible light when electrons return to the ground state.
D.Mobile valence electrons move throughout the metallic structure and absorb light of all frequencies.
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PastPaper.workedSolution
In the presence of ligands, the degenerate d-orbitals split into two groups of different energy levels. When visible light falls on the complex, an electron can be promoted from a lower energy d-orbital to a higher energy d-orbital (d-d transition) by absorbing a specific frequency of visible light. The remaining non-absorbed light (the complementary colour) is transmitted, which is the colour we observe.
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1 mark for the correct option B.
PastPaper.question 15 · multiple-choice
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The standard enthalpy changes of formation, \(\Delta H_f^\theta\), of gaseous butane (\(\text{C}_4\text{H}_{10}\)), carbon dioxide (\(\text{CO}_2\)), and liquid water (\(\text{H}_2\text{O}\)) are given in the table below: \(\text{C}_4\text{H}_{10}(\text{g})\) is \(-126\text{ kJ mol}^{-1}\); \(\text{CO}_2(\text{g})\) is \(-394\text{ kJ mol}^{-1}\); \(\text{H}_2\text{O}(\text{l})\) is \(-286\text{ kJ mol}^{-1}\). What is the standard enthalpy change of combustion of gaseous butane, \(\Delta H_c^\theta[\text{C}_4\text{H}_{10}(\text{g})]\)?
A.-4310 kJ mol^{-1}
B.-2880 kJ mol^{-1}
C.-554 kJ mol^{-1}
D.+2880 kJ mol^{-1}
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PastPaper.workedSolution
The equation for the combustion of butane is: \(\text{C}_4\text{H}_{10}(\text{g}) + 6.5\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 5\text{H}_2\text{O}(\text{l})\). Using Hess's law: \(\Delta H_c^\theta = \sum \Delta H_f^\theta(\text{products}) - \sum \Delta H_f^\theta(\text{reactants}) = [4 \times (-394) + 5 \times (-286)] - [-126] = [-1576 - 1430] + 126 = -2880\text{ kJ mol}^{-1}\).
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1 mark for the correct option B.
PastPaper.question 16 · multiple-choice
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A mixture of \(1.00\text{ mol}\) of \(\text{X}_2(\text{g})\) and \(1.00\text{ mol}\) of \(\text{Y}_2(\text{g})\) is placed in a sealed vessel of volume \(V\text{ dm}^3\) and allowed to reach equilibrium at temperature \(T\). \(\text{X}_2(\text{g}) + \text{Y}_2(\text{g}) \rightleftharpoons 2\text{XY}(\text{g})\). At equilibrium, the concentration of \(\text{XY}(\text{g})\) is \(1.50\text{ mol dm}^{-3}\) and the equilibrium constant \(K_c\) for the reaction is \(9.00\). What is the volume, \(V\), of the vessel?
A.0.50 dm^3
B.0.80 dm^3
C.1.20 dm^3
D.1.50 dm^3
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PastPaper.workedSolution
Let \(x\) moles of \(\text{X}_2\) react to reach equilibrium. The equilibrium moles of \(\text{X}_2\), \(\text{Y}_2\), and \(\text{XY}\) are \(1.00-x\), \(1.00-x\), and \(2x\) respectively. \(K_c = \frac{[\text{XY}]^2}{[\text{X}_2][\text{Y}_2]} = \frac{(2x)^2}{(1.00-x)^2} = 9.00\). Taking the square root of both sides gives: \(\frac{2x}{1.00-x} = 3.00 \Rightarrow 2x = 3.00 - 3.00x \Rightarrow 5.00x = 3.00 \Rightarrow x = 0.60\text{ mol}\). The equilibrium moles of \(\text{XY}\) is \(2x = 1.20\text{ mol}\). Since the equilibrium concentration of \(\text{XY}\) is \(1.50\text{ mol dm}^{-3}\), we have \(\frac{1.20}{V} = 1.50 \Rightarrow V = 0.80\text{ dm}^3\).
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1 mark for the correct option B.
PastPaper.question 17 · multiple-choice
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10\text{ cm}^3 of a gaseous hydrocarbon \(C_xH_y\) is reacted with excess oxygen (100\text{ cm}^3). After complete combustion and cooling to room temperature and pressure, the remaining gas volume is 80\text{ cm}^3. When this remaining gas is passed through aqueous sodium hydroxide, the volume decreases to 40\text{ cm}^3.
What is the molecular formula of the hydrocarbon?
A.\(C_4H_6\)
B.\(C_4H_8\)
C.\(C_4H_{10}\)
D.\(C_3H_8\)
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PastPaper.workedSolution
First, determine the volume of \(CO_2\) produced: the decrease in volume upon passing through aqueous NaOH is due to the absorption of \(CO_2\). Hence, \(\text{Volume of } CO_2 = 80\text{ cm}^3 - 40\text{ cm}^3 = 40\text{ cm}^3\).
Since 10\text{ cm}^3 of \(C_xH_y\) produces 40\text{ cm}^3 of \(CO_2\), \(x = \frac{40}{10} = 4\).
Next, find the volume of \(O_2\) reacted: the remaining gas after NaOH treatment is the unreacted oxygen, which is 40\text{ cm}^3. Since the initial volume of oxygen was 100\text{ cm}^3, the volume of oxygen that reacted is: \(\text{Volume of } O_2 \text{ reacted} = 100\text{ cm}^3 - 40\text{ cm}^3 = 60\text{ cm}^3\).
The combustion equation is: \(C_4H_y + 6O_2 \rightarrow 4CO_2 + \frac{y}{2}H_2O\).
From the stoichiometry of oxygen: \(4 + \frac{y}{4} = 6 \implies \frac{y}{4} = 2 \implies y = 8\).
Therefore, the molecular formula of the hydrocarbon is \(C_4H_8\).
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1 mark for selecting the correct option (B).
PastPaper.question 18 · multiple-choice
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An equimolar mixture of two Period 3 oxides is added to a large excess of water. The resulting mixture is stirred thoroughly and then filtered to remove any insoluble material. The pH of the filtrate is measured at 25 \(^{\circ}\text{C}\) and found to be approximately 2.
Which pair of oxides could have been used?
A.\(Na_2O\) and \(SiO_2\)
B.\(MgO\) and \(Al_2O_3\)
C.\(SiO_2\) and \(P_4O_{10}\)
D.\(Al_2O_3\) and \(SiO_2\)
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PastPaper.workedSolution
\(SiO_2\) is a giant covalent macromolecular oxide that is completely insoluble in water. When the mixture is filtered, \(SiO_2\) remains on the filter paper as a residue.
\(P_4O_{10}\) is a simple molecular acidic oxide. It reacts vigorously with water to form phosphoric(V) acid, \(H_3PO_4\): \(P_4O_{10}(s) + 6H_2O(l) \rightarrow 4H_3PO_4(aq)\).
Since \(H_3PO_4\) is a moderately strong acid in aqueous solution, the filtrate will be strongly acidic with a pH of approximately 2.
For the other options: \(Na_2O\) forms highly alkaline \(NaOH\) (pH \(\approx 13-14\)); \(MgO\) forms a weakly alkaline solution (pH \(\approx 9\)) and \(Al_2O_3\) is insoluble; both \(Al_2O_3\) and \(SiO_2\) are insoluble in water, which would yield a neutral filtrate (pH \(\approx 7\)).
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1 mark for selecting the correct option (C).
PastPaper.question 19 · multiple-choice
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Which statement best explains why aqueous transition metal complexes, such as \([Cu(H_2O)_6]^{2+}\), are coloured?
A.All five d-orbitals remain degenerate, and light is emitted when electrons transition between them.
B.Ligands absorb specific wavelengths of light, promoting ligand electrons to higher energy d-orbitals.
C.The d-orbitals split into two non-degenerate energy levels, and an electron is promoted to a higher level by absorbing a specific frequency of visible light.
D.Energy is released as visible light when d-orbitals undergo splitting in the presence of ligands.
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PastPaper.workedSolution
In a transition metal complex, the presence of ligands causes the five degenerate d-orbitals of the transition metal ion to split into two non-degenerate energy levels with an energy gap, \(\Delta E\). When white light passes through the solution, d-electrons can absorb a photon with energy equal to \(\Delta E\) (\(\Delta E = h\nu\)) to be promoted from the lower d-orbital energy level to the higher level (a d\u2013d transition). The frequency of this absorbed light lies in the visible spectrum, and the complementary colour is transmitted and observed.
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1 mark for selecting the correct option (C).
PastPaper.question 20 · multiple-choice
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A cobalt complex has the formula \([Co(NH_3)_4Cl_2]^+\).
Which row correctly describes the coordination number of cobalt, its oxidation state, and the total number of stereoisomers possible for this complex?
A.Coordination number: 4; Oxidation state: +3; Number of stereoisomers: 2
B.Coordination number: 6; Oxidation state: +3; Number of stereoisomers: 2
C.Coordination number: 6; Oxidation state: +2; Number of stereoisomers: 3
D.Coordination number: 6; Oxidation state: +3; Number of stereoisomers: 4
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PastPaper.workedSolution
1. **Coordination number**: Cobalt is bonded to four \(NH_3\) molecules and two \(Cl^-\)\u200b ions. Since all six ligands are monodentate, the coordination number is 6. 2. **Oxidation state**: Let \(x\) be the oxidation state of cobalt. Neutral ligands (\(NH_3\)) contribute 0, and anionic ligands (\(Cl^-\)) contribute \(-1\) each. The overall charge is \(+1\): \(x + 4(0) + 2(-1) = +1 \implies x = +3\). 3. **Stereoisomers**: The complex is of the type \([MA_4B_2]\) (octahedral geometry), which exhibits geometric isomerism: \(\text{cis}\)-[Co(NH3)4Cl2]+ and \(\text{trans}\)-[Co(NH3)4Cl2]+. Neither geometric isomer contains a chiral centre or lacks a plane of symmetry, so there are no optical isomers. Thus, the total number of stereoisomers is 2.
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PastPaper.question 21 · multiple-choice
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The standard enthalpies of combustion, \(\Delta H_c^{\ominus}\), for carbon, hydrogen, and propanoic acid, \(C_2H_5COOH(l)\), are given below:
The following reversible reaction is allowed to reach equilibrium in a sealed container of fixed volume:
\(2NO_2(g) \rightleftharpoons N_2O_4(g) \quad \Delta H = -57\text{ kJ mol}^{-1}\)
The temperature of the container is increased.
Which statement correctly describes the effect of this change on the position of equilibrium and the value of the equilibrium constant, \(K_c\)?
A.The position of equilibrium shifts to the left, and the value of \(K_c\) decreases.
B.The position of equilibrium shifts to the left, and the value of \(K_c\) remains unchanged.
C.The position of equilibrium shifts to the right, and the value of \(K_c\) increases.
D.The position of equilibrium shifts to the right, and the value of \(K_c\) remains unchanged.
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The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, an increase in temperature shifts the position of equilibrium in the endothermic direction (to the left, in favour of the reactants) to absorb the added thermal energy.
Since the equilibrium constant expression is given by \(K_c = \frac{[N_2O_4]}{[NO_2]^2}\), a shift to the left decreases the concentration of the product (\(N_2O_4\)) and increases the concentration of the reactant (\(NO_2\)). Therefore, the value of \(K_c\) decreases. Note that temperature is the only factor that alters the value of \(K_c\).
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PastPaper.question 23 · multiple-choice
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What is the total number of ions present in 25.0\text{ cm}^3 of 0.120\text{ mol dm}^{-3} aqueous aluminium sulfate, \(Al_2(SO_4)_3\)?
1. Calculate the number of moles of \(Al_2(SO_4)_3\): \(n = C \times V = 0.120\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 0.00300\text{ mol}\).
2. Determine the number of ions per formula unit of \(Al_2(SO_4)_3\): \(Al_2(SO_4)_3(aq) \rightarrow 2Al^{3+}(aq) + 3SO_4^{2-}(aq)\). Each formula unit dissociates to yield 5 ions.
3. Calculate the total moles of ions: \(\text{Moles of ions} = 0.00300\text{ mol} \times 5 = 0.0150\text{ mol}\).
4. Calculate the total number of ions: \(\text{Number of ions} = 0.0150\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 9.03 \times 10^{21}\).
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PastPaper.question 24 · multiple-choice
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Two test-tubes containing water have a small amount of a Period 3 chloride added to them.
- Test-tube 1: Chloride \(X\) dissolves to form a solution with pH \(\approx 7\). - Test-tube 2: Chloride \(Y\) reacts rapidly with water, releasing misty white fumes and forming a solution with pH \(\approx 2\).
Which row correctly identifies Chloride \(X\) and Chloride \(Y\)?
A.Chloride X: \(NaCl\); Chloride Y: \(SiCl_4\)
B.Chloride X: \(MgCl_2\); Chloride Y: \(NaCl\)
C.Chloride X: \(AlCl_3\); Chloride Y: \(SiCl_4\)
D.Chloride X: \(NaCl\); Chloride Y: \(MgCl_2\)
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Chloride \(X\) is \(NaCl\). Sodium chloride is an ionic compound that dissolves in water to give a neutral solution (pH \(\approx 7\)) because the sodium ions and chloride ions undergo simple hydration without any significant hydrolysis.
Chloride \(Y\) is \(SiCl_4\). Silicon tetrachloride is a simple covalent liquid that undergoes rapid hydrolysis in water, producing hydrogen chloride gas (misty white fumes) and a highly acidic solution (pH \(\approx 2\)): \(SiCl_4(l) + 2H_2O(l) \rightarrow SiO_2(s) + 4HCl(aq)\).
Note that \(AlCl_3\) reacts with water to give an acidic solution (pH \(\approx 3\)) but does not react as violently as \(SiCl_4\), and \(MgCl_2\) dissolves to form a weakly acidic solution (pH \(\approx 6.5\)).
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1 mark for selecting the correct option (A).
PastPaper.question 25 · Multiple Choice
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A \(20.0\text{ cm}^3\) sample of a gaseous hydrocarbon, \(\text{C}_x\text{H}_y\), is exploded with an excess of oxygen (\(150.0\text{ cm}^3\)). After cooling to room temperature and pressure, the total volume of gas remaining is \(110.0\text{ cm}^3\). When this remaining gas is shaken with aqueous sodium hydroxide, the volume of gas decreases to \(50.0\text{ cm}^3\).
What is the molecular formula of the hydrocarbon?
A.\(\text{C}_3\text{H}_6\)
B.\(\text{C}_3\text{H}_8\)
C.\(\text{C}_4\text{H}_{10}\)
D.\(\text{C}_2\text{H}_6\)
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1. **Determine the volume of carbon dioxide produced:** Shaking with aqueous \(\text{NaOH}\) removes all of the acidic carbon dioxide gas. \(\text{Volume of CO}_2 = 110.0\text{ cm}^3 - 50.0\text{ cm}^3 = 60.0\text{ cm}^3\). Since 1 mole of \(\text{C}_x\text{H}_y\) produces \(x\) moles of \(\text{CO}_2\), by gaseous volume ratio: \(x = \frac{60.0\text{ cm}^3}{20.0\text{ cm}^3} = 3\).
2. **Determine the volume of oxygen reacted:** The remaining gas after shaking with \(\text{NaOH}\) is the unreacted oxygen (\(50.0\text{ cm}^3\)). \(\text{Volume of O}_2\text{ reacted} = 150.0\text{ cm}^3 - 50.0\text{ cm}^3 = 100.0\text{ cm}^3\).
3. **Use the general combustion equation to find \(y\):** \(\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\) From the volume ratio of \(\text{O}_2\) to hydrocarbon: \(x + \frac{y}{4} = \frac{100.0\text{ cm}^3}{20.0\text{ cm}^3} = 5\) Substitute \(x = 3\): \(3 + \frac{y}{4} = 5 \implies \frac{y}{4} = 2 \implies y = 8\).
Thus, the molecular formula of the hydrocarbon is \(\text{C}_3\text{H}_8\).
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Award 1 mark for the correct option B. - Award 0 marks for incorrect options: - Option A represents propene, which requires less oxygen (4.5 mol per mol). - Option C represents butane, which would produce more carbon dioxide. - Option D represents ethane, which has the wrong carbon-to-hydrogen ratio.
PastPaper.question 26 · Multiple Choice
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An excess of water is added to a solid mixture containing equal moles of silicon(IV) chloride, \(\text{SiCl}_4\), and phosphorus(V) chloride, \(\text{PCl}_5\).
Which statement about the resulting mixture is correct?
A.A completely clear, single-phase solution is formed.
B.A white precipitate is formed, and 9 moles of \(\text{Cl}^-(aq)\) ions are present in the solution for every mole of \(\text{SiCl}_4\) in the starting mixture.
C.No precipitate is formed, and phosphoric(V) acid is the only acid present.
D.A white precipitate is formed, and 4 moles of \(\text{Cl}^-(aq)\) ions are present in the solution for every mole of \(\text{SiCl}_4\) in the starting mixture.
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Both \(\text{SiCl}_4\) and \(\text{PCl}_5\) undergo complete hydrolysis when reacted with excess water: - For silicon(IV) chloride: \(\text{SiCl}_4(l) + 2\text{H}_2\text{O}(l) \rightarrow \text{SiO}_2(s) + 4\text{HCl}(aq)\) - For phosphorus(V) chloride: \(\text{PCl}_5(s) + 4\text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{PO}_4(aq) + 5\text{HCl}(aq)\)
The insoluble \(\text{SiO}_2\) is observed as a white precipitate. Let the initial mixture contain \(1\text{ mol}\) of \(\text{SiCl}_4\) and \(1\text{ mol}\) of \(\text{PCl}_5\). - \(1\text{ mol}\) of \(\text{SiCl}_4\) yields \(4\text{ mol}\) of \(\text{Cl}^-(aq)\). - \(1\text{ mol}\) of \(\text{PCl}_5\) yields \(5\text{ mol}\) of \(\text{Cl}^-(aq)\). Total moles of \(\text{Cl}^-(aq)\) produced per mole of \(\text{SiCl}_4\) is \(4 + 5 = 9\text{ mol}\).
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Award 1 mark for the correct option B. - Option A is incorrect because a white precipitate of silica is formed. - Option C is incorrect because both hydrochloric acid and phosphoric(V) acid are present. - Option D is incorrect because the chloride contribution from both hydrolysed compounds must be summed.
PastPaper.question 27 · Multiple Choice
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The standard enthalpy changes of combustion of carbon (graphite), hydrogen gas, and propan-1-ol are given below:
What is the standard enthalpy change of formation, \(\Delta H^\ominus_f\), of propan-1-ol?
A.\(-302.7\text{ kJ mol}^{-1}\)
B.\(+302.7\text{ kJ mol}^{-1}\)
C.\(-2323.7\text{ kJ mol}^{-1}\)
D.\(-2700.3\text{ kJ mol}^{-1}\)
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The formation equation of propan-1-ol is: \(3\text{C(graphite)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH(l)}\)
Using Hess's Law with standard enthalpy changes of combustion: \(\Delta H^\ominus_f = \sum \Delta H^\ominus_c\text{(reactants)} - \sum \Delta H^\ominus_c\text{(products)}\) \(\Delta H^\ominus_f = [3 \times (-393.5) + 4 \times (-285.8)] - [-2021.0]\) \(\Delta H^\ominus_f = [-1180.5 - 1143.2] + 2021.0\) \(\Delta H^\ominus_f = -2323.7 + 2021.0 = -302.7\text{ kJ mol}^{-1}\)
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Award 1 mark for the correct calculation and option A. - Option B is incorrect due to a sign error during the subtraction. - Option C is the sum of the combustion enthalpies of the constituent elements without accounting for the combustion of propan-1-ol. - Option D represents a simple sum of the three values in the prompt.
PastPaper.question 28 · Multiple Choice
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A mixture of \(1.00\text{ mol}\) of \(\text{X}_2\text{(g)}\) and \(3.00\text{ mol}\) of \(\text{Y}_2\text{(g)}\) is placed in a closed container of volume \(2.00\text{ dm}^3\) and allowed to reach equilibrium at a constant temperature.
Award 1 mark for the correct option B. - Option A is incorrect because the division by volume was omitted. - Option C is incorrect due to failure to apply correct stoichiometric ratios when finding the equilibrium moles of reactants. - Option D is incorrect because the concentrations were not raised to their stoichiometric powers in the \(K_c\) expression.
PastPaper.question 29 · Multiple Choice
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A sample of hydrated copper(II) sulfate is dissolved in water to form a blue solution. An excess of concentrated hydrochloric acid is added dropwise to this solution, resulting in a yellow-green solution.
Which of the following describes the changes in coordination number, geometry, and oxidation state of the copper species during this reaction?
A.Coordination number decreases; geometry changes from octahedral to tetrahedral; oxidation state remains unchanged.
B.Coordination number decreases; geometry changes from octahedral to square planar; oxidation state remains unchanged.
C.Coordination number increases; geometry changes from tetrahedral to octahedral; oxidation state increases.
D.Coordination number decreases; geometry changes from octahedral to tetrahedral; oxidation state decreases.
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The transition is from the octahedral hexaaquacopper(II) complex, \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\), to the tetrahedral tetrachlorocuprate(II) complex, \([\text{CuCl}_4]^{2-}\). - The coordination number decreases from 6 to 4. - The geometry changes from octahedral to tetrahedral. - The oxidation state of copper remains \(+2\) throughout.
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Award 1 mark for the correct option A. - Option B is incorrect because \([\text{CuCl}_4]^{2-}\) is tetrahedral, not square planar, and the oxidation state remains unchanged. - Option C is incorrect because the coordination number decreases. - Option D is incorrect because the copper oxidation state is not reduced.
PastPaper.question 30 · Multiple Choice
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An element \(X\) in Period 3 forms an oxide \(X_2\text{O}_3\) that reacts with both aqueous sodium hydroxide and dilute hydrochloric acid. Another element \(Y\) in Period 3 forms an oxide \(Y\text{O}_2\) that reacts with aqueous sodium hydroxide but not with dilute hydrochloric acid.
Which statement about \(X\) and \(Y\) is correct?
A.Element \(X\) has a higher electronegativity than element \(Y\).
B.The oxide of \(X\) contains ionic bonds with covalent character, whereas the oxide of \(Y\) has a giant covalent structure.
C.The chloride of \(Y\) reacts with water to form a neutral solution.
D.The atomic radius of \(X\) is smaller than that of \(Y\).
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Identify the elements: - Oxide \(X_2\text{O}_3\) is amphoteric, meaning \(X\) is aluminium (\(\text{Al}\)). - Oxide \(Y\text{O}_2\) is acidic and giant covalent, meaning \(Y\) is silicon (\(\text{Si}\)).
Evaluate the statements: - The oxide of \(X\) (\(\text{Al}_2\text{O}_3\)) is ionic with significant covalent character, while the oxide of \(Y\) (\(\text{SiO}_2\)) is a giant covalent structure. This is correct. - Electronegativity increases across Period 3, so \(\text{Si}\) has a higher electronegativity than \(\text{Al}\). Thus, Option A is incorrect. - \(\text{SiCl}_4\) hydrolyses in water to form an acidic solution (due to \(\text{HCl}\)), not a neutral one. Thus, Option C is incorrect. - Atomic radius decreases across Period 3, so \(\text{Al}\) (\(X\)) is larger than \(\text{Si}\) (\(Y\)). Thus, Option D is incorrect.
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Award 1 mark for the correct option B. - Option A reversed the trend of electronegativity. - Option C incorrect because Period 3 non-metal chlorides hydrolyse to produce strongly acidic solutions. - Option D reversed the trend of atomic radius.
PastPaper.question 31 · Multiple Choice
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The reaction between gaseous nitrogen and gaseous hydrogen to form gaseous hydrazine, \(\text{N}_2\text{H}_4\text{(g)}\), is represented by the equation below:
Some average bond energies are given in the table: - \(\text{N}\equiv\text{N} = 945\text{ kJ mol}^{-1}\) - \(\text{H}-\text{H} = 436\text{ kJ mol}^{-1}\) - \(\text{N}-\text{H} = 391\text{ kJ mol}^{-1}\)
What is the average bond energy of the \(\text{N}-\text{N}\) single bond in hydrazine?
A.\(158\text{ kJ mol}^{-1}\)
B.\(243\text{ kJ mol}^{-1}\)
C.\(348\text{ kJ mol}^{-1}\)
D.\(501\text{ kJ mol}^{-1}\)
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Using the relation: \(\Delta H = \sum E_{\text{reactants}} - \sum E_{\text{products}}\)
Set up the equation: \(+95 = 1817 - [E(\text{N}-\text{N}) + 1564]\) \(+95 = 253 - E(\text{N}-\text{N})\) \(E(\text{N}-\text{N}) = 253 - 95 = 158\text{ kJ mol}^{-1}\)
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Award 1 mark for the correct calculation and option A. - Option C is incorrect due to a sign error: adding \(95\) to \(253\) instead of subtracting. - Options B and D represent errors in stoichiometric calculations of the reactant or product bonds.
PastPaper.question 32 · Multiple Choice
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An element \(Z\) has three naturally occurring isotopes. The mass spectrum of \(Z\) shows the following peaks: - \(^{24}Z\) relative abundance \(79.0\%\) - \(^{25}Z\) relative abundance \(10.0\%\) - \(^{26}Z\) relative abundance \(11.0\%\)
A sample of \(Z\) reacts with an excess of dilute hydrochloric acid to produce a salt \(Z\text{Cl}_2\) and hydrogen gas.
What is the mass of \(Z\text{Cl}_2\) produced when \(0.120\text{ mol}\) of \(Z\) reacts completely with hydrochloric acid? [\(A_r(\text{Cl}) = 35.5\)]
A.\(2.92\text{ g}\)
B.\(7.18\text{ g}\)
C.\(11.4\text{ g}\)
D.\(22.9\text{ g}\)
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1. **Calculate the relative atomic mass of \(Z\):** \(A_r(Z) = \frac{(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0)}{100} = \frac{1896 + 250 + 286}{100} = 24.32\)
2. **Identify the molar mass of \(Z\text{Cl}_2\):** \(M_r(Z\text{Cl}_2) = 24.32 + 2(35.5) = 95.32\text{ g mol}^{-1}\)
3. **Calculate the mass of \(0.120\text{ mol}\) of \(Z\text{Cl}_2\):** \(\text{Mass} = n \times M_r = 0.120 \times 95.32 = 11.44\text{ g}\), which rounds to \(11.4\text{ g}\).
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Award 1 mark for the correct calculation and option C. - Option A represents only the mass of reactant metal \(Z\). - Option B is incorrect because only one chlorine atom was added to the formula mass of the salt. - Option D is incorrect because the stoichiometric ratio of \(Z\) to \(Z\text{Cl}_2\) is \(1:1\), not \(1:2\).
PastPaper.question 33 · multiple-choice
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A sample of 1.38 g of an anhydrous metal carbonate, \(X_2\text{CO}_3\), was dissolved in excess dilute hydrochloric acid.
The volume of carbon dioxide gas collected at room temperature and pressure (r.t.p.) was 240 \(\text{cm}^3\).
What is the identity of the metal \(X\)?
[Take the molar volume of a gas at r.t.p. as 24.0 \(\text{dm}^3\text{ mol}^{-1}\).]
A.Lithium
B.Sodium
C.Potassium
D.Rubidium
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1. Calculate the number of moles of \(\text{CO}_2\) produced: \(n(\text{CO}_2) = \frac{240\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0100\text{ mol}\).
2. Determine the moles of \(X_2\text{CO}_3\) reacted: From the balanced equation, 1 mole of \(X_2\text{CO}_3\) produces 1 mole of \(\text{CO}_2\). So, \(n(X_2\text{CO}_3) = 0.0100\text{ mol}\).
3. Calculate the molar mass (\(M_r\)) of \(X_2\text{CO}_3\): \(M_r = \frac{\text{mass}}{\text{moles}} = \frac{1.38\text{ g}}{0.0100\text{ mol}} = 138\text{ g mol}^{-1}\).
The element in Group 1 with a relative atomic mass closest to 39.0 is potassium (\(A_r = 39.1\)).
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1 mark for selecting option C.
PastPaper.question 34 · multiple-choice
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How many total ions are present in a solution prepared by dissolving 9.52 g of anhydrous magnesium chloride, \(\text{MgCl}_2\), in water to make a total volume of 500 \(\text{cm}^3\)?
1. Calculate the molar mass of \(\text{MgCl}_2\): \(M_r(\text{MgCl}_2) = 24.3 + 2(35.5) = 95.3\text{ g mol}^{-1}\).
2. Calculate the moles of \(\text{MgCl}_2\) dissolved: \(n = \frac{9.52\text{ g}}{95.3\text{ g mol}^{-1}} = 0.100\text{ mol}\).
3. Determine the total moles of ions in the solution: Each formula unit of \(\text{MgCl}_2\) dissociates into 3 ions: \(\text{MgCl}_2(\text{s}) \rightarrow \text{Mg}^{2+}(\text{aq}) + 2\text{Cl}^-(\text{aq})\). Total moles of ions \(= 3 \times 0.100\text{ mol} = 0.300\text{ mol}\).
4. Calculate the total number of ions: \(\text{Total ions} = 0.300\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 1.81 \times 10^{23}\).
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1 mark for selecting option C.
PastPaper.question 35 · multiple-choice
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Two Period 3 elements, \(Y\) and \(Z\), each form a chloride that reacts with or dissolves in water.
- The chloride of \(Y\) reacts vigorously with water to form a strongly acidic solution (\(\text{pH} \approx 2\)) and white fumes. - The chloride of \(Z\) dissolves in water to form a neutral solution (\(\text{pH} \approx 7\)).
Which elements are represented by \(Y\) and \(Z\)?
A.\(Y\) is silicon; \(Z\) is phosphorus
B.\(Y\) is phosphorus; \(Z\) is magnesium
C.\(Y\) is silicon; \(Z\) is sodium
D.\(Y\) is sodium; \(Z\) is phosphorus
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PastPaper.workedSolution
- The chloride of silicon, \(\text{SiCl}_4\), reacts vigorously with water (hydrolyses) to produce a strongly acidic solution of hydrochloric acid and white fumes of \(\text{HCl}\) gas: \(\text{SiCl}_4(\text{l}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{SiO}_2(\text{s}) + 4\text{HCl}(\text{g})\). Thus, \(Y\) is silicon. - The chloride of sodium, \(\text{NaCl}\), dissolves in water without hydrolysing to form a neutral solution (\(\text{pH} \approx 7\)). Thus, \(Z\) is sodium. - Therefore, option C is the correct response.
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1 mark for selecting option C.
PastPaper.question 36 · multiple-choice
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An element \(X\) in Period 3 is burned in oxygen to form an oxide \(Y\).
Oxide \(Y\) is a white solid that reacts with both dilute hydrochloric acid and aqueous sodium hydroxide.
What is the identity of element \(X\)?
A.Magnesium
B.Aluminium
C.Silicon
D.Phosphorus
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PastPaper.workedSolution
- Magnesium burns to form \(\text{MgO}\), which is a basic metal oxide that reacts only with acids, not bases. - Aluminium burns to form \(\text{Al}_2\text{O}_3\), which is an amphoteric oxide. It reacts with both acids (forming aluminium salts) and alkalis (forming aluminate ions). - Silicon burns to form \(\text{SiO}_2\), which is a weakly acidic giant covalent oxide that does not react with dilute acids like \(\text{HCl}\). - Phosphorus burns to form \(\text{P}_4\text{O}_{10}\), which is an acidic molecular oxide. - Thus, \(X\) must be aluminium.
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1 mark for selecting option B.
PastPaper.question 37 · multiple-choice
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Which transition metal species has the electronic configuration \([\text{Ar}] 3\text{d}^5\)?
A.\(\text{Fe}^{2+}\)
B.\(\text{Mn}^{2+}\)
C.\(\text{Cr}^{2+}\)
D.\(\text{Co}^{3+}\)
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PastPaper.workedSolution
- Iron (\(\text{Fe}\), \(Z=26\)) has the electronic configuration \([\text{Ar}] 3\text{d}^6 4\text{s}^2\). Removing 2 electrons to form \(\text{Fe}^{2+}\) yields \([\text{Ar}] 3\text{d}^6\). - Manganese (\(\text{Mn}\), \(Z=25\)) has the electronic configuration \([\text{Ar}] 3\text{d}^5 4\text{s}^2\). Removing 2 electrons to form \(\text{Mn}^{2+}\) yields \([\text{Ar}] 3\text{d}^5\). - Chromium (\(\text{Cr}\), \(Z=24\)) has the electronic configuration \([\text{Ar}] 3\text{d}^5 4\text{s}^1\). Removing 2 electrons to form \(\text{Cr}^{2+}\) yields \([\text{Ar}] 3\text{d}^4\). - Cobalt (\(\text{Co}\), \(Z=27\)) has the electronic configuration \([\text{Ar}] 3\text{d}^7 4\text{s}^2\). Removing 3 electrons to form \(\text{Co}^{3+}\) yields \([\text{Ar}] 3\text{d}^6\). - Therefore, \(\text{Mn}^{2+}\) has the correct configuration.
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1 mark for selecting option B.
PastPaper.question 38 · multiple-choice
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Which statement correctly explains the origin of colour in transition metal complexes such as \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\)?
A.Electrons absorb light energy when they are promoted from a lower energy 3d orbital to a higher energy 3d orbital.
B.Electrons emit light energy in the visible spectrum when they drop from a 4s orbital to a 3d orbital.
C.Light is emitted when dative covalent bonds are formed between the ligands and the transition metal ion.
D.The absorption of light is due to the promotion of electrons from the 3d subshell to the 4s subshell.
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PastPaper.workedSolution
In a transition metal complex, the coordinate bonding from ligands causes the five d-orbitals of the transition metal ion to split into two groups of different energy levels. When visible light shines on the complex, an electron in a lower energy 3d orbital absorbs a photon of light and is promoted to a higher energy 3d orbital (d-d transition). The light not absorbed is transmitted or reflected, producing the complementary colour that we observe.
The volume of the container is suddenly halved while keeping the temperature constant.
Which statement about the system immediately after this change is correct?
A.The value of the equilibrium constant, \(K_c\), increases.
B.The rate of the forward reaction increases while the rate of the reverse reaction decreases.
C.The yield of \(\text{SO}_3(\text{g})\) increases because the equilibrium position shifts to the right.
D.The total pressure in the container remains unchanged because the temperature is constant.
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PastPaper.workedSolution
- A: \(K_c\) is temperature-dependent only. Because temperature is constant, \(K_c\) does not change. - B: Halving the volume increases the concentration of all species. Therefore, both forward and reverse reaction rates increase. - C: There are 3 moles of gas on the left and 2 moles of gas on the right. Halving the volume increases the pressure. According to Le Chatelier's principle, the equilibrium position shifts to the right to oppose the pressure increase, resulting in an increased yield of \(\text{SO}_3(\text{g})\). - D: The total pressure in the container increases because the same amount of gas is constrained in half the original volume.
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1 mark for selecting option C.
Paper 21 (AS Level Structured Questions)
Answer all questions. Write your answers in the spaces provided on the question paper.
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PastPaper.question 1 · Structured
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This question is about determining the empirical and molecular formulas of an organic compound and investigating its reactions.
(a) Define the terms: (i) relative molecular mass, \(M_r\) [1] (ii) empirical formula [1] (iii) State the difference between an empirical formula and a molecular formula. [1]
(b) An organic compound \(X\) contains only carbon, hydrogen, and chlorine. Complete combustion of a 1.485 g sample of \(X\) in excess oxygen produced 1.320 g of carbon dioxide, \(CO_2\), and 0.540 g of water, \(H_2O\). Calculate the empirical formula of \(X\). Show your working. [4]
(c) A 0.297 g sample of \(X\) was vaporised in a gas syringe at 373 K and a pressure of \(1.01 \times 10^5\text{ Pa}\). The volume of the vaporised gas was measured as \(92.1\text{ cm}^3\). (i) Use the ideal gas equation, \(pV = nRT\), to calculate the relative molecular mass, \(M_r\), of \(X\). [3] (ii) Deduce the molecular formula of \(X\). [1]
(d) Compound \(X\) is identified as 1,2-dichloroethane. (i) Write a balanced chemical equation for the hydrolysis of 1,2-dichloroethane with excess aqueous sodium hydroxide to form ethane-1,2-diol, \(CH_2(OH)CH_2(OH)\). [1] (ii) In an experiment, 4.95 g of 1,2-dichloroethane is heated with excess aqueous sodium hydroxide. The reaction has a percentage yield of 75.0%. Calculate the mass of ethane-1,2-diol produced. (\(M_r\) of ethane-1,2-diol = 62.0). [3]
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(a) (i) Relative molecular mass, \(M_r\), is the weighted average mass of a molecule of a compound relative to 1/12th of the mass of an atom of carbon-12. (ii) Empirical formula is the simplest whole-number ratio of atoms of each element present in a compound. (iii) The empirical formula shows the simplest whole-number ratio of atoms, whereas the molecular formula shows the actual number of atoms of each element in one molecule of the compound.
(b) - Mass of Carbon = \(1.320 \times \frac{12.0}{44.0} = 0.360\text{ g}\) - Moles of Carbon = \(0.360 / 12.0 = 0.0300\text{ mol}\) - Mass of Hydrogen = \(0.540 \times \frac{2.0}{18.0} = 0.060\text{ g}\) - Moles of Hydrogen = \(0.060 / 1.0 = 0.0600\text{ mol}\) - Mass of Chlorine = \(1.485 - (0.360 + 0.060) = 1.065\text{ g}\) - Moles of Chlorine = \(1.065 / 35.5 = 0.0300\text{ mol}\) - Mole ratio C : H : Cl = \(0.0300 : 0.0600 : 0.0300 = 1 : 2 : 1\) - Empirical formula = \(CH_2Cl\)
(c) (i) \(pV = nRT \implies n = \frac{pV}{RT}\) - Convert volume to \(\text{m}^3\): \(V = 92.1 \times 10^{-6}\text{ m}^3\) - \(n = \frac{1.01 \times 10^5 \times 92.1 \times 10^{-6}}{8.31 \times 373} = \frac{9.3021}{3100} \approx 0.00300\text{ mol}\) - \(M_r = \frac{\text{mass}}{n} = \frac{0.297}{0.00300} = 99.0\) (ii) Empirical formula mass of \(CH_2Cl = 12.0 + (2 \times 1.0) + 35.5 = 49.5\). Since \(99.0 / 49.5 = 2\), the molecular formula is \(C_2H_4Cl_2\).
(d) (i) \(CH_2ClCH_2Cl + 2NaOH \rightarrow CH_2(OH)CH_2(OH) + 2NaCl\) (ii) - Moles of \(C_2H_4Cl_2\) used = \(4.95 / 99.0 = 0.0500\text{ mol}\) - Moles of ethane-1,2-diol expected (1:1 ratio) = 0.0500 mol - Actual moles of product with 75.0% yield = \(0.0500 \times 0.750 = 0.0375\text{ mol}\) - Mass of ethane-1,2-diol = \(0.0375 \times 62.0 = 2.325\text{ g}\) (or 2.33 g)
PastPaper.markingScheme
(a) (i) [1 mark]: Weighted average mass of a molecule relative to 1/12th of the mass of an atom of carbon-12. (ii) [1 mark]: Simplest whole-number ratio of atoms of each element in a compound. (iii) [1 mark]: Empirical is simplest ratio, molecular is the actual number of atoms.
(b) - [1 mark]: Calculates mass and moles of carbon (0.360 g, 0.0300 mol). - [1 mark]: Calculates mass and moles of hydrogen (0.060 g, 0.0600 mol). - [1 mark]: Calculates mass and moles of chlorine by difference (1.065 g, 0.0300 mol). - [1 mark]: Deduces empirical formula CH2Cl with correct simplified ratio working.
(c) - [1 mark]: Uses correct unit conversion for volume \(V = 92.1 \times 10^{-6}\text{ m}^3\). - [1 mark]: Calculates moles \(n = 0.00300\text{ mol}\) using \(pV = nRT\). - [1 mark]: Calculates molecular mass \(M_r = 99.0\). - [1 mark]: Deduces molecular formula \(C_2H_4Cl_2\).
(d) - [1 mark]: Correct balanced equation for hydrolysis with NaOH. - [1 mark]: Calculates starting moles of reactant (0.0500 mol). - [1 mark]: Applies 75% yield and calculates final mass 2.33 g (accept 2.325 g).
PastPaper.question 2 · Structured
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This question concerns the chemical periodicity of the elements in Period 3 of the Periodic Table.
(a) Sodium and sulfur both react with oxygen when heated. (i) Describe the observations made during each reaction, and write a balanced equation for each reaction. [4] (ii) State the type of bonding and structure shown by sodium oxide and sulfur dioxide. [2]
(b) The melting points of silicon dioxide, \(SiO_2\), and phosphorus(V) oxide, \(P_4O_{10}\), are shown below: - Melting point of \(SiO_2\) = 1883 K - Melting point of \(P_4O_{10}\) = 573 K Explain this difference in melting points in terms of structure and bonding. [4]
(c) Phosphorus(V) chloride, \(PCl_5\), is added to water. (i) Describe what is observed and write a balanced chemical equation for the reaction. [3] (ii) Predict the approximate pH of the resulting solution and explain your prediction. [2]
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(a) (i) - Sodium: Burns with a yellow/orange flame to form a white solid. - Equation: \(4Na(s) + O_2(g) \rightarrow 2Na_2O(s)\) (or \(2Na(s) + O_2(g) \rightarrow Na_2O_2(s)\)) - Sulfur: Burns with a blue flame to produce a colorless gas with a pungent, choking smell. - Equation: \(S(s) + O_2(g) \rightarrow SO_2(g)\) (ii) - Sodium oxide: Ionic bonding and giant ionic structure. - Sulfur dioxide: Covalent bonding and simple molecular structure.
(b) - Silicon dioxide has a giant covalent macromolecular structure. - Melting silicon dioxide requires breaking many strong covalent bonds throughout the 3D lattice, which requires a very large amount of thermal energy. - Phosphorus(V) oxide has a simple molecular structure. - Melting phosphorus(V) oxide only requires overcoming weak intermolecular forces (London dispersion forces / van der Waals forces) between molecules, which requires much less energy.
(c) (i) - Observation: Vigorous/violent reaction, white/misty fumes (of \(HCl\) gas) are evolved, and a colorless solution is formed. - Equation: \(PCl_5(s) + 4H_2O(l) \rightarrow H_3PO_4(aq) + 5HCl(aq)\) (or \(PCl_5(s) + H_2O(l) \rightarrow POCl_3(l) + 2HCl(aq)\)) (ii) - pH: 1 or 2 (strongly acidic). - Explanation: Highly acidic hydrogen ions (\(H^+\)) are released from the complete dissociation of \(HCl\) in water.
PastPaper.markingScheme
(a) - [1 mark]: Correct description of yellow flame and white solid for sodium, and blue flame for sulfur. - [1 mark]: Correct balanced equation for reaction of sodium with oxygen. - [1 mark]: Correct balanced equation for reaction of sulfur with oxygen. - [1 mark]: Deduces correct types of bonding (ionic for sodium oxide, covalent for sulfur dioxide). - [1 mark]: Deduces correct structures (giant ionic for sodium oxide, simple molecular for sulfur dioxide).
(b) - [1 mark]: Mentions that \(SiO_2\) has a giant covalent macromolecular structure. - [1 mark]: Explains that melting \(SiO_2\) requires breaking strong covalent bonds. - [1 mark]: Mentions that \(P_4O_{10}\) has a simple molecular structure. - [1 mark]: Explains that melting \(P_4O_{10}\) only requires overcoming weak intermolecular forces (not covalent bonds).
(c) - [1 mark]: Describes observations: vigorous reaction and misty/white fumes. - [1 mark]: Correct balanced equation: \(PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl\). - [1 mark]: Predicts pH of 1 or 2. - [1 mark]: Explains that \(HCl\) is a strong acid that dissociates fully to produce \(H^+\) ions.
PastPaper.question 3 · Structured
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This question is about chemical energetics, calorimetry, and Hess's Law.
(a) A student determines the enthalpy change of neutralisation, \(\Delta H_{neut}\), of hydrochloric acid with aqueous sodium hydroxide. \(50.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\ HCl(aq)\) is mixed with \(50.0\text{ cm}^3\) of \(1.05\text{ mol dm}^{-3}\ NaOH(aq)\) in a polystyrene cup. Both solutions are initially at 21.5 °C. The maximum temperature reached is 28.2 °C. Assume the density of the solution is \(1.00\text{ g cm}^{-3}\) and its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\). (i) Calculate the heat energy, \(q\), in Joules, released during this reaction. [2] (ii) Show that \(HCl\) is the limiting reagent and state the number of moles of water formed. [2] (iii) Calculate the enthalpy change of neutralisation, \(\Delta H_{neut}\), in \(\text{kJ mol}^{-1}\), for this reaction. Include a sign. [2]
(b) Explain why the experimental value of \(\Delta H_{neut}\) obtained is less exothermic than the standard literature value of \(-57.1\text{ kJ mol}^{-1}\), and suggest a practical improvement to the setup. [2]
(c) Standard enthalpy changes of combustion, \(\Delta H_c^\ominus\), are given below: - \(\Delta H_c^\ominus [C(s)] = -393.5\text{ kJ mol}^{-1}\) - \(\Delta H_c^\ominus [H_2(g)] = -285.8\text{ kJ mol}^{-1}\) - \(\Delta H_c^\ominus [C_2H_5COOH(l)] = -1527.0\text{ kJ mol}^{-1}\)
(i) Write the chemical equation, with state symbols, for the standard enthalpy change of formation of liquid propanoic acid, \(C_2H_5COOH\). [2] (ii) Draw a labeled Hess’s Law cycle connecting the formation of propanoic acid with its combustion products. [2] (iii) Calculate the standard enthalpy change of formation, \(\Delta H_f^\ominus\), of liquid propanoic acid in \(\text{kJ mol}^{-1}\). [3]
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(a) (i) - Total mass of solution = \(50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\). - Temperature rise, \(\Delta T = 28.2 - 21.5 = 6.7\text{ K}\). - \(q = mc\Delta T = 100.0 \times 4.18 \times 6.7 = 2800.6\text{ J}\). (ii) - Moles of \(HCl = 0.0500\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0500\text{ mol}\). - Moles of \(NaOH = 0.0500\text{ dm}^3 \times 1.05\text{ mol dm}^{-3} = 0.0525\text{ mol}\). - Since 1 mole of \(HCl\) reacts with 1 mole of \(NaOH\), \(HCl\) is in a smaller molar amount and is the limiting reagent. - Moles of water formed = 0.0500 mol. (iii) - \(q = 2.8006\text{ kJ}\). - \(\Delta H_{neut} = -\frac{q}{n} = -\frac{2.8006}{0.0500} = -56.012\text{ kJ mol}^{-1}\). - To 3 sig figs: \(-56.0\text{ kJ mol}^{-1}\).
(b) - Reason: Heat is lost to the surroundings (or absorbed by the polystyrene cup / thermometer). - Improvement: Use a lid on the polystyrene cup (or wrap the cup in an insulating material like cotton wool).
(c) (i) \(3C(s) + 3H_2(g) + O_2(g) \rightarrow C_2H_5COOH(l)\) (ii) The Hess's Law cycle shows: - Elements: \(3C(s) + 3H_2(g) + O_2(g)\) at the top/left. - Propanoic acid: \(C_2H_5COOH(l)\) on the right. - Combustion Products: \(3CO_2(g) + 3H_2O(l)\) at the bottom. - Arrow from Elements to Propanoic acid: \(\Delta H_f^\ominus\). - Arrow from Elements to Combustion Products: \(3\Delta H_c^\ominus[C] + 3\Delta H_c^\ominus[H_2]\). - Arrow from Propanoic acid to Combustion Products: \(\Delta H_c^\ominus[C_2H_5COOH]\). (iii) - \(\Delta H_f^\ominus = 3\Delta H_c^\ominus[C] + 3\Delta H_c^\ominus[H_2] - \Delta H_c^\ominus[C_2H_5COOH]\) - \(\Delta H_f^\ominus = 3(-393.5) + 3(-285.8) - (-1527.0)\) - \(\Delta H_f^\ominus = -1180.5 - 857.4 + 1527.0 = -2037.9 + 1527.0 = -510.9\text{ kJ mol}^{-1}\).
PastPaper.markingScheme
(a) - [1 mark]: Correct substitution into \(q = mc\Delta T\) using mass of 100.0 g. - [1 mark]: Calculates heat energy \(q = 2800.6\text{ J}\) (or 2.80 kJ). - [1 mark]: Shows moles of \(HCl\) and \(NaOH\) and identifies \(HCl\) as limiting. - [1 mark]: Deduces moles of water formed = 0.0500 mol. - [1 mark]: Divides heat energy by moles of water. - [1 mark]: Correctly quotes \(-56.0\text{ kJ mol}^{-1}\) with negative sign and appropriate units.
(b) - [1 mark]: Heat lost to surroundings. - [1 mark]: Practical improvement (e.g., add a lid / insulate cup).
(c) - [1 mark]: Writes correct balanced equation for formation of propanoic acid. - [1 mark]: Includes all state symbols correctly (\(C(s)\), \(H_2(g)\), \(O_2(g)\), \(C_2H_5COOH(l)\)). - [2 marks]: Draws complete Hess's Law cycle with correct labels and arrow directions. - [1 mark]: Correct expression of Hess's Law pathway. - [1 mark]: Substitution of correct values. - [1 mark]: Calculates \(-510.9\text{ kJ mol}^{-1}\) with correct sign and units.
PastPaper.question 4 · Structured
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This question is about chemical equilibria and the determination of equilibrium constants.
Ethanoic acid reacts with ethanol in a reversible reaction to form ethyl ethanoate and water. The equation for this reaction is: \[CH_3COOH(l) + C_2H_5OH(l) \rightleftharpoons CH_3COOC_2H_5(l) + H_2O(l)\]
(a) Write the expression for the equilibrium constant, \(K_c\), for this reaction. State why the volumes of the mixture cancel in the expression for \(K_c\). [2]
(b) In an experiment, 1.00 mol of ethanoic acid and 1.50 mol of ethanol are mixed in a sealed container and allowed to reach equilibrium at a constant temperature, \(T\). At equilibrium, it is found that 0.80 mol of ethyl ethanoate has been formed. Calculate the value of \(K_c\) at temperature \(T\). Show your working. [4]
(c) A different equilibrium mixture is prepared at the same temperature, \(T\), containing 2.00 mol of ethanoic acid, 1.00 mol of ethanol, and 0.50 mol of water. Let \(x\) represent the equilibrium amount, in moles, of ethyl ethanoate. Write an expression using your calculated value of \(K_c\) from (b) to show how \(x\) could be calculated. (You do not need to solve the equation). [3]
(d) Use Le Chatelier's principle to predict and explain the effect on the position of equilibrium and the value of \(K_c\) when: (i) More ethanol is added to the mixture. [3] (ii) The temperature is increased, given that the forward reaction is exothermic. [3]
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(a) - \(K_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}\) - The volumes cancel out because there are the same number of moles of reactants and products (2 moles on each side). The volume term, \(V\), appears to the power of 2 in both the numerator and the denominator, so they cancel completely.
Substitute the equilibrium amounts into the \(K_c\) expression: - \(K_c = \frac{0.80 \times 0.80}{0.20 \times 0.70} = \frac{0.64}{0.14} \approx 4.57\) (no units as they cancel out).
(c) Set up the ICE table for the new mixture: - \(CH_3COOH\): Equilibrium = \(2.00 - x\) mol - \(C_2H_5OH\): Equilibrium = \(1.00 - x\) mol - \(CH_3COOC_2H_5\): Equilibrium = \(x\) mol - \(H_2O\): Equilibrium = \(0.50 + x\) mol
Substitute into the \(K_c\) expression: - \(K_c = \frac{x(0.50 + x)}{(2.00 - x)(1.00 - x)} = 4.57\)
(d) (i) - Position of equilibrium: Shifts to the right (the product side). - Explanation: To oppose the increase in concentration of ethanol by consuming it. - Value of \(K_c\): Remains unchanged because \(K_c\) is only affected by temperature. (ii) - Position of equilibrium: Shifts to the left (the reactant side). - Explanation: Since the forward reaction is exothermic, increasing temperature shifts the equilibrium in the endothermic (reverse) direction to absorb the added heat. - Value of \(K_c\): Decreases because the concentration of products at equilibrium decreases while the concentration of reactants increases.
PastPaper.markingScheme
(a) - [1 mark]: Correctly writes the \(K_c\) expression. - [1 mark]: Explains that equal number of moles on both sides causes the volume terms to cancel out.
(b) - [1 mark]: Calculates equilibrium moles of \(CH_3COOH = 0.20\text{ mol}\). - [1 mark]: Calculates equilibrium moles of \(C_2H_5OH = 0.70\text{ mol}\). - [1 mark]: Identifies equilibrium moles of \(H_2O = 0.80\text{ mol}\). - [1 mark]: Correctly calculates \(K_c = 4.57\) (accept 4.6, no units required).
(c) - [1 mark]: Identifies equilibrium concentrations in terms of \(x\) (especially \(0.50+x\) for water). - [1 mark]: Correctly sets up the fraction with \(x\). - [1 mark]: Equates the fraction to the calculated value of \(K_c\) (e.g. 4.57).
(d) - [1 mark]: Deduces equilibrium shifts right when ethanol is added, with explanation. - [1 mark]: States \(K_c\) is constant when concentration changes. - [1 mark]: Deduces equilibrium shifts left when temperature increases, with explanation (forward is exothermic). - [1 mark]: States \(K_c\) decreases when temperature increases for an exothermic reaction.
Paper 31 (Advanced Practical Skills)
Answer all questions. Show your working and appropriate significant figures in final calculations.
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PastPaper.question 1 · practical
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### Practical Task 1: Enthalpy Change of Neutralisation
In this experiment, you are to determine the enthalpy change of neutralisation, \(\Delta H_{\text{neut}}\), when a weak monobasic acid, \(\text{HA}\), is neutralised by aqueous sodium hydroxide, \(\text{NaOH}\).
1. A student transferred \(25.0\text{ cm}^3\) of \(1.60\text{ mol dm}^{-3}\) aqueous sodium hydroxide, \(\text{NaOH}\), into a polystyrene cup. 2. The student measured the initial temperature of the \(\text{NaOH}\) solution. 3. A second student transferred \(25.0\text{ cm}^3\) of \(1.60\text{ mol dm}^{-3}\) weak acid, \(\text{HA}\), into a beaker and measured its temperature, which was identical to that of the \(\text{NaOH}\). 4. The student then added the acid quickly to the polystyrene cup, stirred the mixture continuously, and recorded the maximum temperature reached.
**Results** - Initial temperature of both solutions = \(21.4\text{ }^\circ\text{C}\) - Maximum temperature reached = \(31.2\text{ }^\circ\text{C}\)
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### Questions
**(a)** Calculate the temperature rise, \(\Delta T\), of the mixture.
$$\Delta T = \text{................... }^\circ\text{C}$$
[1]
**(b)** Calculate the heat energy, \(q\), released in the reaction in joules. *(Assume that the density of the reaction mixture is \(1.00\text{ g cm}^{-3}\) and that its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\).)*
$$q = \text{................... J}$$
[2]
**(c)** Calculate the number of moles of water formed during this neutralisation reaction.
$$\text{moles of H}_2\text{O} = \text{................... mol}$$
[2]
**(d)** Use your answers to **(b)** and **(c)** to calculate the enthalpy change of neutralisation, \(\Delta H_{\text{neut}}\), in \(\text{kJ mol}^{-1}\). Include the appropriate sign and give your answer to **three** significant figures.
**(e)** In an improved experiment, a student records the temperature at \(1\)-minute intervals for both the solutions and the mixture, and uses a cooling curve extrapolation. Explain why this extrapolation method provides a more accurate value for the temperature change than simply recording the maximum temperature.
[2]
**(f)** Sketch the graph of temperature against time that would be obtained for this experiment. - Clearly label the axes. - Assume the solutions are mixed at \(t = 3\text{ min}\). - Show how the extrapolation is performed to find the theoretical maximum temperature rise, \(\Delta T_{\text{theory}}\).
[3]
**(g)** Another student carried out the exact same experiment but used a simple glass beaker instead of a polystyrene cup. State and explain how the temperature rise and the magnitude of the calculated \(\Delta H_{\text{neut}}\) would compare to those obtained using a polystyrene cup.
[2]
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PastPaper.workedSolution
**(a)** $$\Delta T = 31.2 - 21.4 = 9.8\text{ }^\circ\text{C}$$
**(b)** Total volume of solution = \(25.0 + 25.0 = 50.0\text{ cm}^3\) Since density is \(1.00\text{ g cm}^{-3}\), the mass, \(m = 50.0\text{ g}\). $$q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 9.8\text{ K} = 2048.2\text{ J}$$ *(Accept \(2048\text{ J}\) or \(2050\text{ J}\))*
**(c)** $$\text{Moles of HA} = \frac{25.0}{1000} \times 1.60 = 0.0400\text{ mol}$$ $$\text{Moles of NaOH} = \frac{25.0}{1000} \times 1.60 = 0.0400\text{ mol}$$ Since the acid is monobasic and reacts in a 1:1 ratio, the moles of \(\text{H}_2\text{O}\) formed is \(0.0400\text{ mol}\).
**(d)** First convert \(q\) to kilojoules: $$q = 2.0482\text{ kJ}$$ $$\Delta H_{\text{neut}} = -\frac{q}{n} = -\frac{2.0482\text{ kJ}}{0.0400\text{ mol}} = -51.205\text{ kJ mol}^{-1}$$ To 3 significant figures and with the correct negative sign: $$\Delta H_{\text{neut}} = -51.2\text{ kJ mol}^{-1}$$
**(e)** Heat loss to the surroundings occurs from the moment the reaction starts. Extrapolating the cooling curve back to the time of mixing (3 minutes) corrects for this heat loss, allowing the determination of the theoretical maximum temperature change that would have occurred if the reaction had been instantaneous and no heat had been lost.
**(f)** The sketched graph should feature: 1. Temperature on the y-axis and Time on the x-axis. 2. A horizontal line showing a steady initial temperature until \(t = 3\text{ min}\), followed by a sharp rise, and then a line curving gradually downwards after the peak. 3. A vertical extrapolation line drawn at \(t = 3\text{ min}\) that meets the backward-extrapolated cooling curve line, with \(\Delta T\) clearly indicated as the vertical distance between the initial temperature and the extrapolated peak.
**(g)** Both the recorded temperature rise and the magnitude of the calculated \(\Delta H_{\text{neut}}\) would be lower (less exothermic). Glass is a much better conductor of heat (a poorer insulator) than polystyrene, so a significant amount of heat energy would escape to the surroundings during the reaction, resulting in a lower maximum temperature.
PastPaper.markingScheme
**Part (a)** - M1: Award 1 mark for the correct calculation of the temperature rise: \(9.8\text{ }^\circ\text{C}\).
**Part (b)** - M1: Award 1 mark for identifying the correct mass of the mixture as \(50.0\text{ g}\). - M2: Award 1 mark for the correct evaluation of heat energy: \(2048.2\text{ J}\) (or \(2048\text{ J}\) / \(2050\text{ J}\)).
**Part (c)** - M1: Award 1 mark for showing correct working: \(\frac{25.0}{1000} \times 1.60\). - M2: Award 1 mark for calculating \(0.0400\text{ mol}\) (must be written to at least 3 significant figures).
**Part (d)** - M1: Award 1 mark for dividing the heat in kJ (derived in part b) by the number of moles (derived in part c). - M2: Award 1 mark for including the correct negative sign and unit (\(\text{kJ mol}^{-1}\)). - M3: Award 1 mark for calculating the final value to 3 significant figures (\(-51.2\) or error carried forward from candidate's previous values).
**Part (e)** - M1: Award 1 mark for noting that heat is lost to the surroundings as the reaction proceeds. - M2: Award 1 mark for explaining that extrapolation estimates the maximum temperature that would be reached in the absence of heat loss (or if the reaction were instantaneous).
**Part (f)** - M1: Award 1 mark for correctly labelled axes: Temperature / \(T\) (including units, \(^\circ\text{C}\)) on the y-axis and Time / \(t\) (including units, \(\text{min}\)) on the x-axis. - M2: Award 1 mark for a curve showing a constant temperature up to \(t = 3\text{ min}\), a rapid rise, and then a gradual linear or near-linear decrease. - M3: Award 1 mark for showing a vertical line at \(t = 3\text{ min}\) and the backward extrapolation of the cooling curve to this line to find the maximum theoretical temperature.
**Part (g)** - M1: Award 1 mark for stating that both the temperature rise and the calculated magnitude of \(\Delta H_{\text{neut}}\) would be lower (or less negative / less exothermic). - M2: Award 1 mark for explaining that glass is a better thermal conductor than polystyrene (or polystyrene is a better insulator), resulting in more heat being lost to the surroundings.
PastPaper.question 2 · practical
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An experiment was carried out to determine the enthalpy change of hydration of anhydrous magnesium sulfate, \(\Delta H_{\text{hyd}}\), which cannot be measured directly.
Two experiments were conducted using a simple polystyrene cup calorimeter. [Assume the specific heat capacity of the solution is \(4.18 \text{ J g}^{-1} \text{ K}^{-1}\) and the density of the solution is \(1.00 \text{ g cm}^{-3}\).]
**Experiment 1: Enthalpy change of solution of anhydrous magnesium sulfate, \(\Delta H_1\)** * Mass of weighing bottle + \(\text{MgSO}_4\text{(s)}\) = \(14.85 \text{ g}\) * Mass of empty weighing bottle = \(9.15 \text{ g}\) * Volume of distilled water in the plastic cup = \(50.0 \text{ cm}^3\) * Initial temperature of water = \(21.5 \ ^\circ\text{C}\) * Maximum temperature reached = \(39.0 \ ^\circ\text{C}\)
**Experiment 2: Enthalpy change of solution of hydrated magnesium sulfate, \(\Delta H_2\)** * Mass of weighing bottle + \(\text{MgSO}_4 \cdot 7\text{H}_2\text{O(s)}\) = \(21.95 \text{ g}\) * Mass of empty weighing bottle = \(10.25 \text{ g}\) * Volume of distilled water in the plastic cup = \(50.0 \text{ cm}^3\) * Initial temperature of water = \(22.0 \ ^\circ\text{C}\) * Minimum temperature reached = \(18.5 \ ^\circ\text{C}\)
**(a)** Calculate the heat energy released, \(q_1\), in Joules, during Experiment 1.
**(b)** Calculate the enthalpy change of solution of anhydrous magnesium sulfate, \(\Delta H_1\), in \(\text{kJ mol}^{-1}\). Show your working and give your answer to 3 significant figures. \([A_{\text{r}}: \text{Mg} = 24.3, \text{S} = 32.1, \text{O} = 16.0, \text{H} = 1.0]\)
**(c)** Calculate the enthalpy change of solution of hydrated magnesium sulfate, \(\Delta H_2\), in \(\text{kJ mol}^{-1}\). Show your working and give your answer to 3 significant figures.
**(d)** Construct an energy cycle using Hess's Law to connect the three enthalpy changes, and calculate the enthalpy change of hydration, \(\Delta H_{\text{hyd}}\), of anhydrous magnesium sulfate in \(\text{kJ mol}^{-1}\).
**(e)** (i) Calculate the percentage uncertainty in the temperature change, \(\Delta T\), for Experiment 1, if the thermometer used has an uncertainty of \(\pm 0.5 \ ^\circ\text{C}\) for each individual reading. Show your working. (ii) Suggest one improvement to the experimental setup to reduce heat loss, other than adding a lid.
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**(a)** \(q_1 = m c \Delta T\) \(m = 50.0 \text{ g}\) (assuming density of water is \(1.00 \text{ g cm}^{-3}\)) \(\Delta T_1 = 39.0 - 21.5 = 17.5 \ ^\circ\text{C}\) \(q_1 = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 17.5 \text{ K} = 3657.5 \text{ J}\) (or \(3660 \text{ J}\))
**(b)** Mass of \(\text{MgSO}_4\) used = \(14.85 - 9.15 = 5.70 \text{ g}\) \(M_{\text{r}}(\text{MgSO}_4) = 24.3 + 32.1 + 4(16.0) = 120.4 \text{ g mol}^{-1}\) \(n(\text{MgSO}_4) = \frac{5.70}{120.4} = 0.04734 \text{ mol}\) \(\Delta H_1 = -\frac{q_1}{1000 \times n} = -\frac{3.6575 \text{ kJ}}{0.04734 \text{ mol}} = -77.3 \text{ kJ mol}^{-1}\) (to 3 sig figs, with negative sign for exothermic reaction)
**(d)** Using the Hess's Law cycle: \(\text{MgSO}_4\text{(s)} + 7\text{H}_2\text{O(l)} \xrightarrow{\Delta H_{\text{hyd}}} \text{MgSO}_4 \cdot 7\text{H}_2\text{O(s)}\) Both dissolve to form the same aqueous solution of \(\text{MgSO}_4\text{(aq)}\). Therefore, \(\Delta H_1 = \Delta H_{\text{hyd}} + \Delta H_2\) \(\Delta H_{\text{hyd}} = \Delta H_1 - \Delta H_2\) \(\Delta H_{\text{hyd}} = -77.3 - (+15.4) = -92.7 \text{ kJ mol}^{-1}\) (using unrounded intermediate values gives \(-77.26 - 15.41 = -92.7 \text{ kJ mol}^{-1}\))
**(e)** (i) For a temperature change, there are two thermometer readings (initial and maximum/minimum). Total absolute uncertainty = \(2 \times 0.5 \ ^\circ\text{C} = 1.0 \ ^\circ\text{C}\) Percentage uncertainty = \(\frac{1.0}{17.5} \times 100\% = 5.71\%\) (accept \(5.7\%\)) (ii) Place the polystyrene cup inside a larger beaker packed with cotton wool / mineral wool for insulation, or use a double polystyrene cup.
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**(a)** [1 Mark] * Award 1 mark for correct calculation of heat released: \(q_1 = 50.0 \times 4.18 \times 17.5 = 3657.5 \text{ J}\) (or \(3.66 \text{ kJ}\)).
**(b)** [2 Marks] * Award 1 mark for calculating the correct number of moles of \(\text{MgSO}_4\): \(n = 0.0473 \text{ mol}\). * Award 1 mark for calculating \(\Delta H_1 = -77.3 \text{ kJ mol}^{-1}\) (must include the negative sign, units, and be to 3 sig figs).
**(c)** [2 Marks] * Award 1 mark for calculating the correct number of moles of \(\text{MgSO}_4 \cdot 7\text{H}_2\text{O}\): \(n = 0.0475 \text{ mol}\). * Award 1 mark for calculating \(\Delta H_2 = +15.4 \text{ kJ mol}^{-1}\) (must include the positive sign, units, and be to 3 sig figs).
**(d)** [2 Marks] * Award 1 mark for a correct Hess's Law expression: \(\Delta H_{\text{hyd}} = \Delta H_1 - \Delta H_2\). * Award 1 mark for the correct final calculation of \(\Delta H_{\text{hyd}} = -92.7 \text{ kJ mol}^{-1}\) (allow error carried forward from parts b and c).
**(e)** [3 Marks] * (i) Award 1 mark for recognizing that two readings are taken, giving a total uncertainty of \(1.0 \ ^\circ\text{C}\). * (i) Award 1 mark for correct percentage uncertainty: \(\frac{1.0}{17.5} \times 100\% = 5.7\%\). * (ii) Award 1 mark for a valid practical suggestion: e.g., placing the polystyrene cup inside a beaker packed with cotton wool/insulating material, or nesting a second polystyrene cup inside the first (double-cupping).
PastPaper.question 3 · practical
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### Qualitative Analysis
In this task, you will identify the cations and anions present in three aqueous solutions: **FA 5**, **FA 6**, and **FA 7**. Each solution contains **one** cation and **one** anion from the standard list of ions in the Qualitative Analysis Notes.
**(a)** Carry out the following tests and record all your observations in a single, clearly structured table.
* **Test 1:** Add aqueous sodium hydroxide, \(\text{NaOH}\), dropwise until in excess to separate 1 cm depths of **FA 5**, **FA 6**, and **FA 7** in test-tubes. * **Test 2:** Add aqueous ammonia, \(\text{NH}_3\), dropwise until in excess to separate 1 cm depths of **FA 5**, **FA 6**, and **FA 7** in test-tubes. * **Test 3:** Add a few drops of aqueous silver nitrate, \(\text{AgNO}_3\), to separate 1 cm depths of **FA 5**, **FA 6**, and **FA 7** in test-tubes. To any precipitate formed, add aqueous ammonia until in excess. * **Test 4:** Mix a 1 cm depth of **FA 6** with a 1 cm depth of **FA 7** in a test-tube. To the resulting mixture, add dilute nitric acid, \(\text{HNO}_3\).
[8 marks]
**(b)** Use your observations from **(a)** to identify the ions present in **FA 5**, **FA 6**, and **FA 7**. State the evidence for each ion.
* Cation in **FA 5**: * Anion in **FA 5**: * Cation in **FA 6**: * Anion in **FA 6**: * Cation in **FA 7**: * Anion in **FA 7**:
Support each of your identifications with a brief summary of the evidence from your observations.
[5 marks]
**(c)** (i) A student suggests that the cation in **FA 7** could be zinc, \(\text{Zn}^{2+}\), rather than aluminium, \(\text{Al}^{3+}\). Explain how your observations in **(a)** show that **FA 7** contains \(\text{Al}^{3+}\) and not \(\text{Zn}^{2+}\).
(ii) Suggest a chemical test, other than mixing **FA 6** and **FA 7**, that can be used to confirm that the anion in **FA 7** is sulfate, \(\text{SO}_4^{2-}\). State the reagent(s) you would use and the expected observation.
[2 marks]
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### Detailed Solution
#### Part (a): Observations
| Test | Observation for **FA 5** (\(\text{MgI}_2\)) | Observation for **FA 6** (\(\text{BaCl}_2\)) | Observation for **FA 7** (\(\text{Al}_2(\text{SO}_4)_3\)) | | :--- | :--- | :--- | :--- | | **Test 1** (add \(\text{NaOH}\)) | White precipitate, insoluble in excess | No precipitate / no reaction | White precipitate, soluble in excess to form a colorless solution | | **Test 2** (add \(\text{NH}_3\)) | White precipitate, insoluble in excess | No precipitate / no reaction | White precipitate, insoluble in excess | | **Test 3** (add \(\text{AgNO}_3\)) | Yellow precipitate, insoluble in aqueous ammonia | White precipitate, soluble in aqueous ammonia to form a colorless solution | No precipitate / no reaction | | **Test 4** (mix **FA 6** + **FA 7**, then add \(\text{HNO}_3\)) | *N/A* | White precipitate forms on mixing **FA 6** and **FA 7**. The precipitate remains / does not dissolve on adding dilute \(\text{HNO}_3\). | *N/A* |
#### Part (b): Identification and Evidence * **FA 5**: * Cation: \(\text{Mg}^{2+}\). Evidence: White precipitate with both \(\text{NaOH}\) and \(\text{NH}_3\), both of which are insoluble in excess reagent. * Anion: \(\text{I}^-\). Evidence: Yellow precipitate with \(\text{AgNO}_3\) that is insoluble in aqueous ammonia. * **FA 6**: * Cation: \(\text{Ba}^{2+}\). Evidence: No precipitate with either \(\text{NaOH}\) or \(\text{NH}_3\), but forms a white precipitate with sulfate ions in **FA 7** that is insoluble in acid. * Anion: \(\text{Cl}^-\). Evidence: White precipitate with \(\text{AgNO}_3\) that dissolves in aqueous ammonia to give a colorless solution. * **FA 7**: * Cation: \(\text{Al}^{3+}\). Evidence: White precipitate with \(\text{NaOH}\) that is soluble in excess, and a white precipitate with \(\text{NH}_3\) that is insoluble in excess. * Anion: \(\text{SO}_4^{2-}\). Evidence: Forms a white precipitate when mixed with barium ions in **FA 6** which does not dissolve in dilute nitric acid.
#### Part (c): * (i) If zinc ions, \(\text{Zn}^{2+}\), were present, the white precipitate formed with aqueous ammonia would dissolve in excess to give a colorless solution. Because the white precipitate with **FA 7** remained insoluble in excess aqueous ammonia, the cation must be \(\text{Al}^{3+}\). * (ii) Reagents: Add aqueous barium nitrate, \(\text{Ba(NO}_3)_2\), followed by dilute nitric acid, \(\text{HNO}_3\) (or aqueous barium chloride acidified with dilute hydrochloric acid). Observation: A white precipitate forms that remains insoluble in the acid.
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### Marking Scheme
#### Part (a) [8 marks] * **M1 (Table Structure):** Award 1 mark for constructing a single, comprehensive table with clear columns/rows and headings, recording observations for all tests. * **M2 (FA 5 with NaOH & NH3):** Award 1 mark for observing a white precipitate, insoluble in excess for both reagents. * **M3 (FA 6 with NaOH & NH3):** Award 1 mark for observing "no precipitate" or "no change" for both reagents. * **M4 (FA 7 with NaOH & NH3):** Award 1 mark for observing a white precipitate soluble in excess \(\text{NaOH}\) to form a colorless solution AND a white precipitate insoluble in excess \(\text{NH}_3\). * **M5 (Silver Nitrate colors):** Award 1 mark for observing a yellow precipitate with **FA 5**, a white precipitate with **FA 6**, and no precipitate with **FA 7**. * **M6 (Silver Nitrate solubility in NH3):** Award 1 mark for showing that the yellow precipitate (**FA 5**) is insoluble in ammonia AND the white precipitate (**FA 6**) is soluble in ammonia. * **M7 (Mixing FA 6 and FA 7):** Award 1 mark for observing a white precipitate upon mixing **FA 6** and **FA 7**. * **M8 (Effect of Acid on Mixture):** Award 1 mark for observing that the white precipitate remains/does not dissolve on adding dilute \(\text{HNO}_3\).
#### Part (b) [5 marks] * **M9 (Cation Identification):** Award 2 marks for all three correct, or 1 mark for any two correct: * **FA 5:** \(\text{Mg}^{2+}\) * **FA 6:** \(\text{Ba}^{2+}\) * **FA 7:** \(\text{Al}^{3+}\) * **M10 (Anion Identification):** Award 2 marks for all three correct, or 1 mark for any two correct: * **FA 5:** \(\text{I}^-\) * **FA 6:** \(\text{Cl}^-\) * **FA 7:** \(\text{SO}_4^{2-}\) * **M11 (Evidence):** Award 1 mark for providing correct, matching observations to support at least one cation and one anion (e.g., \(\text{Al}^{3+}\) forms a white ppt soluble in excess \(\text{NaOH}\); \(\text{I}^-\) forms a yellow ppt insoluble in \(\text{NH}_3\)).
#### Part (c) [2 marks] * **M12 (Distinguishing Cations):** Award 1 mark for stating that \(\text{Zn}^{2+}\) precipitate would dissolve in excess aqueous ammonia, whereas \(\text{Al}^{3+}\) precipitate is insoluble in excess aqueous ammonia. * **M13 (Sulfate Test):** Award 1 mark for stating the correct reagents: aqueous barium nitrate acidified with dilute nitric acid (or aqueous barium chloride acidified with dilute hydrochloric acid) AND the observation: white precipitate remains/insoluble in acid. *(Reject: Ba2+(aq) alone without acid; Reject: sulfuric acid as the acidifier)*