2024 CCEA A-Level Chemistry 1110 模拟试题及答案详解

\( \Delta G = \Delta H - T\Delta S = -58.0 \times 10^3 - 298 \times (-176) = -58000 + 52448 = -5552 \text{ J mol}^{-1} = -5.55 \text{ kJ mol}^{-1} \). Since \( \Delta G < 0 \), the reaction is feasible at 298 K.

1 mark: A. Distractors B–D arise from sign errors when combining the \( \Delta H \) and \( T\Delta S \) terms.

For a first-order reaction, \( t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{0.693}{4.62 \times 10^{-3}} = 150 \text{ s} \).

1 mark: B. C is double the correct value (dividing by \( 2k \) in error); D omits the \( \ln 2 \) factor.

The Maxwell–Boltzmann distribution shifts to higher energies as T increases, so the proportion of molecules with energy \( \geq E_a \) rises sharply (the area under the curve beyond \( E_a \) increases), giving a much larger \( k \) via \( k = Ae^{-E_a/RT} \). \( E_a \) and A are constants for a given reaction and do not change with temperature.

1 mark: A. B and C misidentify which terms in the Arrhenius equation are temperature-independent.

\( p(NO_2) = 0.60 \times 200 = 120 \text{ kPa} \); \( p(N_2O_4) = 0.40 \times 200 = 80 \text{ kPa} \). \( K_p = \dfrac{p(NO_2)^2}{p(N_2O_4)} = \dfrac{120^2}{80} = \dfrac{14400}{80} = 180 \text{ kPa} \).

1 mark: A. B omits squaring \( p(NO_2) \); D inverts the expression.

\( pK_a = -\log(1.8 \times 10^{-5}) = 4.74 \). \( pH = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} = 4.74 + \log\dfrac{0.30}{0.20} = 4.74 + 0.18 = 4.92 \).

1 mark: A. D is \( pK_a \) alone with the ratio term omitted.

Each \( -COOH \) group has both a hydrogen-bond donor (\( O{-}H \)) and acceptor (\( C{=}O \)). Two ethanoic acid molecules align so that each \( O{-}H \) hydrogen-bonds to the \( C{=}O \) of the other, forming a stable 8-membered cyclic dimer held by two hydrogen bonds.

1 mark: A.

\( RCOOH + SOCl_2 \rightarrow RCOCl + SO_2 + HCl \). Both HCl and \( SO_2 \) gases are evolved, distinguishing \( SOCl_2 \) from \( PCl_5 \), which gives HCl and \( POCl_3 \) (no \( SO_2 \)).

1 mark: A. B is a classic distractor since \( PCl_5 \) also gives steamy fumes, but not \( SO_2 \).

Acid hydrolysis is catalytic and reaches a position of equilibrium (reversible), regenerating the carboxylic acid. Alkaline hydrolysis (saponification) is effectively irreversible because the ethanoate ion formed is a poor electrophile and does not re-esterify, driving the reaction to completion.

1 mark: A.

\( H_2SO_4 \) protonates \( HNO_3 \), which then loses water: \( HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^- \). The nitronium ion \( NO_2^+ \) is the electrophile that is attacked by the delocalised \( \pi \) electrons of the benzene ring.

1 mark: A.

The nitrogen lone pair is delocalised into the aromatic \( \pi \) system (as in phenol), increasing electron density at the 2, 4 and 6 positions and strongly activating the ring, so no halogen-carrier catalyst is needed and substitution occurs at all three positions.

1 mark: A.

\( \Delta S^{\ominus} = \Sigma S^{\ominus}(\text{products}) - \Sigma S^{\ominus}(\text{reactants}) = (39.7 + 213.6) - 92.9 = 253.3 - 92.9 = +160.4 \text{ J K}^{-1}\text{mol}^{-1} \).

1 mark: sum of product entropies (253.3); 1 mark: correct subtraction of reactant entropy; 1 mark: correct final answer with sign and units \( +160.4 \text{ J K}^{-1}\text{mol}^{-1} \). [3]

\( \Delta G = \Delta H - T\Delta S \). Although \( \Delta H \) is positive (unfavourable), \( \Delta S \) is also positive, so the \( -T\Delta S \) term becomes increasingly negative as T rises; above a sufficiently high temperature this term outweighs the positive \( \Delta H \), making \( \Delta G \) negative overall. The reaction is feasible when \( \Delta G \leq 0 \), i.e. at the boundary \( \Delta H = T\Delta S \), so \( T = \dfrac{\Delta H}{\Delta S} = \dfrac{125000}{231} = 541 \text{ K} \).

1 mark: explanation that the \( -T\Delta S \) term becomes more negative/dominant as T increases; 1 mark: condition for feasibility stated as \( \Delta H = T\Delta S \) (i.e. \( \Delta G = 0 \)); 1 mark: correct rearrangement \( T = \Delta H / \Delta S \) with consistent units (125000 J); 1 mark: correct final answer 541 K (accept 540–542 K). [4]

At the temperature limit of feasibility, \( \Delta G = 0 \), so \( \Delta H = T\Delta S = 462 \times (-85) = -39270 \text{ J mol}^{-1} = -39.3 \text{ kJ mol}^{-1} \).

1 mark: recognising \( \Delta G = 0 \) at the limiting temperature; 1 mark: correct substitution \( \Delta H = T\Delta S \); 1 mark: correct final answer \( -39.3 \text{ kJ mol}^{-1} \) (accept \( \pm 0.2 \)). [3]

Comparing experiments 1 and 2: [A] doubles at constant [B], rate increases by a factor of 4 (\( 2^2 \)), so order with respect to A = 2. Comparing experiments 2 and 3: [B] doubles at constant [A], rate doubles, so order with respect to B = 1. Rate equation: \( \text{rate} = k[A]^2[B] \). Using experiment 1: \( k = \dfrac{2.0 \times 10^{-3}}{(0.10)^2(0.10)} = \dfrac{2.0\times10^{-3}}{1.0\times10^{-3}} = 2.0 \text{ mol}^{-2}\text{dm}^6\text{s}^{-1} \).

1 mark: order 2 w.r.t. A, correctly justified from experiments 1–2; 1 mark: order 1 w.r.t. B, correctly justified from experiments 2–3; 1 mark: correct rate equation \( rate = k[A]^2[B] \); 1 mark: correct k with correct units \( \text{mol}^{-2}\text{dm}^6\text{s}^{-1} \) (ECF from stated orders). [4]

\( \ln\dfrac{k_2}{k_1} = -\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right) \). \( \ln\left(\dfrac{1.4\times10^{-2}}{3.5\times10^{-4}}\right) = \ln(40) = 3.689 \). \( \dfrac{1}{340}-\dfrac{1}{300} = -3.922\times10^{-4}\text{ K}^{-1} \). So \( 3.689 = \dfrac{E_a}{8.31}\times 3.922\times10^{-4} \), giving \( E_a = \dfrac{3.689 \times 8.31}{3.922\times10^{-4}} = 78200 \text{ J mol}^{-1} = 78.2 \text{ kJ mol}^{-1} \).

1 mark: correct Arrhenius two-point equation quoted/used; 1 mark: correct evaluation of \( \ln(k_2/k_1) = 3.69 \); 1 mark: correct evaluation of \( (1/T_2 - 1/T_1) \); 1 mark: correct final \( E_a = 78.2 \text{ kJ mol}^{-1} \) (accept 77–79). [4]

The rate of an overall reaction is governed by the slowest step (the rate-determining step). Since step 1 is slow and involves one molecule each of X and Y, the predicted rate equation is \( rate = k[X][Y] \) (first order in each, second order overall). This could be confirmed by the initial-rates method: carrying out a series of experiments in which [X] is varied while [Y] is held constant (and vice versa), measuring the initial rate in each case, and showing that rate is directly proportional to [X] and to [Y] individually.

1 mark: rate equation \( rate=k[X][Y] \) correctly derived from the slow step; 1 mark: correct reasoning that only species in the rate-determining step appear; 1 mark: valid experimental method (initial rates, varying one concentration at a time) to confirm the orders. [3]

Initial \( [H_2] = [I_2] = 1.00/2.00 = 0.500 \text{ mol dm}^{-3} \). Let y = concentration of \( H_2 \) (and \( I_2 \)) that reacts. At equilibrium: \( [H_2]=[I_2]=0.500-y \), \( [HI]=2y \). \( K_c = \dfrac{(2y)^2}{(0.500-y)^2} = 45.0 \). Taking the square root: \( \dfrac{2y}{0.500-y} = \sqrt{45.0} = 6.71 \). Solving: \( 2y = 3.354 - 6.71y \Rightarrow 8.71y = 3.354 \Rightarrow y = 0.385 \text{ mol dm}^{-3} \). \( [HI] = 2y = 0.770 \text{ mol dm}^{-3} \).

1 mark: correct ICE set-up with equilibrium concentrations in terms of y; 1 mark: correct \( K_c \) expression; 1 mark: correct square-root simplification and rearrangement; 1 mark: correct final \( [HI] = 0.770 \text{ mol dm}^{-3} \) (accept 0.76–0.78). [4]

\( [OH^-] = \sqrt{K_b \times c} = \sqrt{1.8\times10^{-5} \times 0.0250} = \sqrt{4.5\times10^{-7}} = 6.71\times10^{-4} \text{ mol dm}^{-3} \). \( pOH = -\log(6.71\times10^{-4}) = 3.17 \). \( pH = 14.00 - pOH = 14.00 - 3.17 = 10.83 \).

1 mark: correct expression \( [OH^-]=\sqrt{K_bc} \); 1 mark: correct \( [OH^-] \) value; 1 mark: correct pOH; 1 mark: correct final pH = 10.83 (accept 10.8). [4]

\( pH = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} \); \( pK_a = -\log(1.8\times10^{-5}) = 4.74 \). \( 4.00 = 4.74 + \log\dfrac{[\text{salt}]}{[\text{acid}]} \Rightarrow \log(\text{ratio}) = -0.74 \Rightarrow \text{ratio} = 10^{-0.74} = 0.180 \). \( [\text{salt}] = 0.180 \times 0.100 = 0.0180 \text{ mol dm}^{-3} \). Moles in 250 cm³ \( = 0.0180 \times 0.250 = 4.50\times10^{-3} \text{ mol} \). Mass \( = 4.50\times10^{-3} \times 82.0 = 0.369 \text{ g} \).

1 mark: correct Henderson–Hasselbalch rearrangement to find the ratio; 1 mark: correct [salt]; 1 mark: correct moles of sodium ethanoate in 250 cm³; 1 mark: correct final mass 0.369 g (accept 0.36–0.38). [4]

Phenolphthalein is the suitable indicator. For a weak acid–strong base titration, the equivalence point occurs above pH 7 (typically around pH 8.5–9) because the ethanoate ion formed is a weak base and hydrolyses slightly in water, making the solution alkaline at the exact equivalence point. The pH curve rises steeply (near-vertically) through roughly pH 7–10 around the equivalence point; this steep region includes phenolphthalein's range (8.3–10.0) but lies above methyl orange's range (3.1–4.4), so methyl orange would change colour too early, in the buffer region before the equivalence point, giving an inaccurate end point.

1 mark: phenolphthalein correctly identified; 1 mark: correct reason that the equivalence point is above pH 7 (ethanoate ion hydrolysis); 1 mark: description of the steep pH jump spanning phenolphthalein's but not methyl orange's range. [3]

Butanoic acid: \( CH_3CH_2CH_2COOH \). Methyl propanoate: \( CH_3CH_2COOCH_3 \). Both compounds share the molecular formula \( C_4H_8O_2 \) but contain different functional groups — a carboxylic acid group (\( -COOH \)) in butanoic acid and an ester group (\( -COO- \)) in methyl propanoate — so they are functional group isomers, and consequently have different chemical properties (e.g. only butanoic acid reacts with carbonates to release \( CO_2 \)).

1 mark: correct structural formula of butanoic acid; 1 mark: correct structural formula of methyl propanoate; 1 mark: 'functional group isomerism' correctly named with a valid reason (same molecular formula, different functional groups). [3]

\( [H^+] = 10^{-2.89} = 1.29\times10^{-3} \text{ mol dm}^{-3} \). Assuming dissociation is small so that \( [HA]_{eqm} \approx 0.0500 \text{ mol dm}^{-3} \): \( K_a = \dfrac{[H^+]^2}{[HA]} = \dfrac{(1.29\times10^{-3})^2}{0.0500} = \dfrac{1.66\times10^{-6}}{0.0500} = 3.32\times10^{-5} \text{ mol dm}^{-3} \).

1 mark: correct \( [H^+] \) from pH; 1 mark: correct \( K_a \) expression \( [H^+]^2/[HA] \); 1 mark: correct substitution; 1 mark: correct final \( K_a \) with units (accept 3.2–3.4 × 10⁻⁵), and assumption that dissociation is negligible compared with initial concentration stated. [4]

Add a spatula of solid sodium hydrogencarbonate (or a few cm³ of aqueous sodium carbonate) to separate samples of each liquid. Propanoic acid, being sufficiently acidic, reacts to give brisk effervescence as carbon dioxide gas is released (\( 2CH_3CH_2COOH + Na_2CO_3 \rightarrow 2CH_3CH_2COONa + H_2O + CO_2 \)); this gas turns limewater milky/cloudy. Propan-1-ol is a neutral alcohol and shows no reaction/no effervescence with the carbonate.

1 mark: correct reagent (sodium hydrogencarbonate/carbonate); 1 mark: correct observation for propanoic acid (effervescence/CO₂ evolved, confirmed with limewater); 1 mark: contrasting observation for propan-1-ol (no reaction). [3]

Propan-1-ol is heated under reflux with an excess of acidified potassium dichromate(VI), \( K_2Cr_2O_7/H_2SO_4(aq) \); the orange dichromate is reduced to green \( Cr^{3+} \) as the alcohol is oxidised. Equation: \( CH_3CH_2CH_2OH + 2[O] \rightarrow CH_3CH_2COOH + H_2O \). Reflux (a condenser returns any evaporated, volatile substances to the flask) is essential because the reaction proceeds via the volatile aldehyde intermediate, propanal; under reflux the propanal cannot escape and remains in contact with the oxidising agent, allowing it to be oxidised further to the acid. If the mixture were simply heated (or set up for distillation), the propanal would evaporate/distil off before being fully oxidised, so the acid would not form.

1 mark: correct reagent/conditions (excess acidified \( K_2Cr_2O_7 \), reflux) and colour change orange to green; 1 mark: correct equation using [O]; 1 mark: propanal identified as the volatile intermediate; 1 mark: correct explanation that reflux retains/returns the intermediate for further oxidation, unlike distillation. [4]

Moles of ethanoic acid \( = \dfrac{5.00}{60.0} = 0.0833 \text{ mol} \). Since the esterification is 1:1, theoretical moles of ester = 0.0833 mol, giving a theoretical mass \( = 0.0833 \times 88.0 = 7.33 \text{ g} \). Actual mass \( = 7.33 \times 0.680 = 4.99 \text{ g} \).

1 mark: correct moles of ethanoic acid; 1 mark: correct theoretical mass of ester (7.33 g); 1 mark: correct application of 68.0% yield; 1 mark: correct final mass 4.99 g (accept 4.9–5.0). [4]

\( CH_3CH_2CONH_2 + NaOH \rightarrow CH_3CH_2COONa + NH_3 \). Ammonia gas, \( NH_3 \), is released; it can be identified because it turns damp red litmus paper blue.

1 mark: correct balanced equation with sodium propanoate and ammonia as products; 1 mark: ammonia correctly identified as the gas; 1 mark: correct confirmatory test (damp red litmus turns blue). [3]

\( (CH_3CO)_2O + C_6H_5OH \rightarrow CH_3COOC_6H_5 + CH_3COOH \). Ethanoic anhydride is generally preferred because it reacts less violently than ethanoyl chloride, does not release fuming, corrosive HCl gas, and is cheaper and less readily hydrolysed by atmospheric moisture, giving a cleaner reaction with a better yield of the ester.

1 mark: correct equation with phenyl ethanoate and ethanoic acid as products; 1 mark: valid advantage (no HCl fumes/less vigorous/cheaper/less moisture-sensitive); 1 mark: brief justification linking the advantage to practical benefit (safer/higher/cleaner yield). [3]

The carbonyl carbon of \( CH_3COCl \) is strongly electrophilic because it is bonded to two electronegative atoms (O and Cl). A lone pair on the oxygen atom of a water molecule attacks this electrophilic carbon, forming a tetrahedral intermediate in which the oxygen now bears a positive charge and the carbonyl oxygen a negative charge. This intermediate is unstable and collapses: the \( C=O \) double bond reforms as the chloride ion is expelled as a leaving group, and a proton is subsequently transferred away from the protonated oxygen (to \( Cl^- \) or another water molecule). Overall: \( CH_3COCl + H_2O \rightarrow CH_3COOH + HCl \); the reaction is vigorous, producing steamy white fumes of HCl.

1 mark: nucleophilic attack of water oxygen lone pair on the electrophilic carbonyl carbon; 1 mark: tetrahedral intermediate correctly described; 1 mark: collapse of intermediate with \( Cl^- \) expelled as \( C=O \) reforms; 1 mark: correct overall equation and observation (steamy white HCl fumes). [4]

\( CH_3COCl + AlCl_3 \rightarrow CH_3CO^+ + AlCl_4^- \). The acylium ion, \( CH_3CO^+ \), is the electrophile generated, which is then attacked by the delocalised \( \pi \) electrons of the benzene ring; \( AlCl_4^- \) subsequently removes a proton from the resulting arenium (Wheland) intermediate to restore aromaticity and regenerate the \( AlCl_3 \) catalyst. The reaction at the ring is electrophilic substitution, specifically Friedel–Crafts acylation.

1 mark: correct equation for electrophile formation; 1 mark: acylium ion \( CH_3CO^+ \) correctly identified as the electrophile; 1 mark: \( AlCl_4^- \) counter-ion/catalyst regeneration role noted; 1 mark: reaction correctly named as electrophilic substitution (Friedel–Crafts acylation). [4]

Predicted enthalpy of hydrogenation of the Kekulé structure \( = 3 \times (-120) = -360 \text{ kJ mol}^{-1} \). Difference \( = -208 - (-360) = +152 \text{ kJ mol}^{-1} \); benzene releases 152 kJ mol⁻¹ less energy than predicted, showing that real benzene is 152 kJ mol⁻¹ lower in energy (more stable/thermodynamically stabilised) than the Kekulé model predicts. This difference is the delocalisation (resonance/aromatic stabilisation) energy, providing evidence that the six \( \pi \) electrons are delocalised equally over all six carbon atoms rather than existing as three localised, alternating double bonds.

1 mark: correct predicted value \( -360 \text{ kJ mol}^{-1} \); 1 mark: correct difference \( +152 \text{ kJ mol}^{-1} \); 1 mark: correct statement that benzene is more stable/lower in energy than predicted; 1 mark: difference identified as the delocalisation/resonance energy, evidencing delocalised \( \pi \) electrons. [4]

(a) Benzoic acid, \( C_6H_5COOH \), is formed. (b) The strong oxidising agent attacks the side-chain specifically at the benzylic carbon (the carbon directly attached to the ring); the C–C bond(s) further along the chain are broken and the remaining carbon atoms are lost as carbon dioxide (and water), leaving only a \( -COOH \) group directly bonded to the ring. Because oxidation always proceeds to this same point regardless of how long the original chain was, any alkylbenzene possessing at least one benzylic C–H is oxidised all the way to benzoic acid.

1 mark: correct product, benzoic acid/\( C_6H_5COOH \); 1 mark: benzylic carbon correctly identified as the site of attack; 1 mark: remainder of the chain described as broken off/lost as \( CO_2 \); 1 mark: correct general statement that any chain length with a benzylic H gives the same product. [4]

(a) The methyl group is 2,4-directing (ortho/para-directing). An alkyl group is electron-donating by induction (the \( +I \) effect), which pushes electron density into the delocalised \( \pi \) system of the ring; this raises electron density most at the carbon atoms ortho and para to the methyl group, making these positions most attractive to attack by the electrophilic \( NO_2^+ \). (b) Under controlled, low-temperature mono-nitration conditions the major product is 4-nitromethylbenzene (1-methyl-4-nitrobenzene, the para isomer), with a smaller amount of the 2-nitro (ortho) isomer also formed; the para product predominates because it is less sterically hindered than the ortho isomer.

1 mark: correctly identified as 2,4-directing (ortho/para-directing); 1 mark: correct explanation via the +I/electron-donating effect of the alkyl group raising ring electron density; 1 mark: correct major product named as 4-nitromethylbenzene/para isomer; 1 mark: valid reasoning that para predominates over ortho (less steric hindrance). [4]

The colorimeter is first set to a wavelength strongly absorbed by aqueous iodine, and calibrated using a series of standard iodine solutions of known concentration to produce a calibration curve of absorbance against \( [I_2] \). Known volumes of propanone, dilute sulfuric acid and iodine solution are mixed, a stopclock is started immediately, and a sample of the mixture is placed in a cuvette in the colorimeter; absorbance readings are taken at regular time intervals and converted into \( [I_2] \) using the calibration curve. A graph of \( [I_2] \) against time is plotted: because this graph is a straight line with a constant negative gradient (rather than a curve which flattens as \( [I_2] \) falls), the rate of reaction is independent of \( [I_2] \), showing the reaction is zero order with respect to iodine. To find the order with respect to propanone, the experiment is repeated several times, each time varying the initial concentration of propanone while keeping \( [I_2] \) and \( [H^+] \) constant; the initial rate (the constant gradient of each \( [I_2] \)–time graph) is measured for each run. A graph of initial rate against initial \( [\text{propanone}] \) is then plotted: a straight line through the origin shows that the reaction is first order with respect to propanone.

Band A (5–6 marks): full, coherent method including wavelength selection and calibration curve, timed absorbance readings converted to \( [I_2] \), the straight-line \( [I_2] \)–time graph as evidence for zero order in \( I_2 \), and a valid method (varying initial [propanone], other concentrations constant, using initial rate) to establish the order in propanone; fluent use of specialist terms, essentially free of errors in spelling, punctuation and grammar. Band B (3–4 marks): most stages present but less complete or less clearly sequenced; reasonable use of specialist terms. Band C (1–2 marks): only a fragmented description, e.g. mentions the colorimeter only; limited use of specialist terms. Indicative content: colorimeter/wavelength selection; calibration curve; timed absorbance readings converted to concentration; straight-line \( [I_2] \)-time graph = zero order in \( I_2 \); vary initial [propanone], other variables constant; use initial rate vs [propanone] graph to deduce first order. [6]

A known mass of 2-hydroxybenzoic acid is placed in a dry flask with an excess of ethanoic anhydride and a few drops of concentrated phosphoric(V) or sulfuric acid catalyst, and the mixture is heated (e.g. in a warm water bath at about 50–60 °C, or under reflux) for around 15 minutes to allow esterification of the phenolic \( -OH \) group to go to completion. The warm mixture is then poured into a beaker of cold water, which hydrolyses any unreacted ethanoic anhydride, and the mixture is cooled in an ice bath to encourage crystallisation of crude aspirin. The crude solid is collected by filtration under reduced pressure (Büchner funnel), washed with a little cold water to remove soluble impurities, and air-dried. To purify by recrystallisation, the crude solid is dissolved in the minimum volume of a hot solvent in which it is soluble hot but only sparingly soluble cold; the hot solution is filtered if necessary (through a fluted, pre-warmed filter paper/funnel) to remove insoluble impurities, and then allowed to cool slowly to room temperature and further in ice, so that pure crystals form while soluble impurities remain dissolved in the mother liquor. The pure crystals are collected by filtration under reduced pressure, washed with a small volume of ice-cold solvent, and dried, for example in a desiccator. The dry mass obtained is compared with the theoretical mass (calculated from the limiting reagent, salicylic acid) to give the percentage yield, and purity is checked by melting point: a sharp melting point close to the literature value (135 °C) indicates a pure product, whereas a lowered or broadened melting range indicates residual impurity.

Band A (5–6 marks): a full, logically sequenced account covering reagents/catalyst and heating conditions, quenching in cold water to hydrolyse excess anhydride, filtration and washing of the crude product, the recrystallisation procedure (minimum hot solvent, optional hot filtration, slow cooling, filtration, washing, drying), and both a yield calculation and a melting-point purity check; fluent use of specialist terms, essentially free from errors in spelling, punctuation and grammar. Band B (3–4 marks): most stages present but sequencing or detail incomplete (e.g. recrystallisation description lacks hot filtration or slow cooling). Band C (1–2 marks): only a fragment of the method described, e.g. mixing reagents and filtering only, with little reference to purification or purity checks. Indicative content: salicylic acid + ethanoic anhydride + acid catalyst, heat; quench in cold water; filter/wash crude solid; recrystallise from minimum hot solvent; hot filtration of insoluble impurity; slow cooling to crystallise; filter, wash, dry; percentage yield from limiting reagent; melting point as purity check. [6]

Moles \( HCl = 0.02240 \times 0.105 = 2.352\times10^{-3} \text{ mol} \). Since \( NaOH + HCl \rightarrow NaCl + H_2O \) is 1:1, moles \( NaOH \) = \( 2.352\times10^{-3} \text{ mol} \). \( [NaOH] = \dfrac{2.352\times10^{-3}}{0.0250} = 0.0941 \text{ mol dm}^{-3} \).

1 mark: A. B halves the moles of acid in error; C doubles it.

Moles \( S_2O_3^{2-} = 0.02460 \times 0.100 = 2.460\times10^{-3} \text{ mol} \). Moles \( I_2 = \tfrac{1}{2}(2.460\times10^{-3}) = 1.230\times10^{-3} \text{ mol} \). From the first equation, moles \( Cu^{2+} = 2 \times \) moles \( I_2 = 2.460\times10^{-3} \text{ mol} \). \( [Cu^{2+}] = \dfrac{2.460\times10^{-3}}{0.0250} = 0.0984 \text{ mol dm}^{-3} \).

1 mark: A. This is a classic two-stage iodometric back-titration; the two stoichiometric factors of 2 cancel.

Moles EDTA \( = 0.01875 \times 0.0200 = 3.75\times10^{-4} \text{ mol} = \) moles \( Ca^{2+} \) (1:1). \( [Ca^{2+}] = \dfrac{3.75\times10^{-4}}{0.0250} = 0.0150 \text{ mol dm}^{-3} \). Mass concentration \( = 0.0150 \times 40.1 = 0.602 \text{ g dm}^{-3} = 602 \text{ mg dm}^{-3} \).

1 mark: A. B halves the correct answer; D doubles it.

Transition metal atoms lose the \( 4s \) electrons first on ionisation, then \( 3d \) electrons as required. \( Fe \rightarrow Fe^{3+} \) loses 3 electrons: both \( 4s \) electrons and one \( 3d \) electron, giving \( [Ar]3d^5 \), a stable half-filled \( 3d \) subshell.

1 mark: A. B is the configuration of \( Fe^{2+} \) (only the two 4s electrons removed).

The larger chloride ligands cannot pack six around the smaller \( Cu^{2+} \) ion as water can, so a ligand-exchange occurs in which six \( H_2O \) ligands are replaced by four \( Cl^- \) ligands, forming the yellow-green tetrahedral complex \( [CuCl_4]^{2-} \); the coordination number falls from 6 to 4 and the geometry changes from octahedral to tetrahedral.

1 mark: A.

By the \( n+1 \) rule, the \( CH_2 \) protons are split into a quartet by the 3 equivalent adjacent \( CH_3 \) protons. Rapid intermolecular proton exchange of the OH proton (with trace water or other alcohol/OH molecules) means it does not maintain a fixed spin relationship with neighbouring protons, so it neither shows resolved splitting itself (appearing as a broad singlet) nor splits the \( CH_2 \) signal.

1 mark: A.

\( E^{\ominus}_{cell} = E^{\ominus}(\text{reduction, more positive}) - E^{\ominus}(\text{oxidation, more negative}) = 0.34 - (-0.76) = +1.10 \text{ V} \). The half-cell with the more positive \( E^{\ominus} \) (Cu²⁺/Cu) is reduced and forms the positive electrode (cathode).

1 mark: A.

In phenylamine, the nitrogen lone pair is delocalised into the aromatic ring, making it far less available to accept a proton, so phenylamine is a weaker base than ammonia. In ethylamine, the electron-donating alkyl group increases electron density on nitrogen (relative to ammonia), making the lone pair more available and ethylamine a stronger base than ammonia. Overall: phenylamine < ammonia < ethylamine.

1 mark: A.

Nitrobenzene is reduced by refluxing with tin and excess concentrated hydrochloric acid, which first forms a soluble tin/anilinium chloride salt complex; excess \( NaOH(aq) \) is then added to liberate and free the phenylamine base, which can be separated (e.g. by steam distillation).

1 mark: A. B is incomplete since the amine remains protonated/complexed until basified; D is an oxidising agent, not a reductant.

At the isoelectric point (a characteristic pH for each amino acid), the amino acid carries no net charge because the number of protonated amine groups (\( -NH_3^+ \)) exactly balances the number of deprotonated carboxylate groups (\( -COO^- \)); it exists as a dipolar ion, the zwitterion, rather than as a neutral un-ionised molecule.

1 mark: A.

(a) \( [CH_3COCH_2CH_2CH_3]^{+\bullet} \rightarrow CH_3CO^+ + {}^{\bullet}CH_2CH_2CH_3 \); the fragment at m/z 43 is the acylium ion \( CH_3CO^+ \). (b) This cleavage is favoured because the resulting acylium cation is stabilised by delocalisation of a lone pair from the oxygen atom into the empty orbital on the positively charged carbon, lowering the energy of this fragmentation pathway relative to other C–C bond cleavages. (c) The M+1 peak appears at m/z = 87; it arises because a small proportion of molecules contain one atom of the naturally occurring \( ^{13}C \) isotope in place of \( ^{12}C \), giving a molecular ion one mass unit heavier.

1 mark: correct fragment ion \( CH_3CO^+ \) and equation; 1 mark: correct reasoning that the acylium ion is stabilised by oxygen lone-pair delocalisation; 1 mark: correct M+1 value (87); 1 mark: correct explanation via the \( ^{13}C \) isotope. [4]

Propan-2-ol has 3 \( ^1H \) environments: the two \( CH_3 \) groups (6H), the \( CH \) (1H) and the \( OH \) (1H), giving peaks in the ratio 6:1:1. The two methyl groups are chemically equivalent because the molecule possesses a plane of symmetry passing through the \( C{-}O{-}H \) and central \( C{-}H \) bonds; both methyl groups therefore experience an identical chemical environment and resonate at the same chemical shift, giving a single combined signal integrating for 6H rather than two separate 3H signals.

1 mark: 3 environments identified; 1 mark: correct ratio 6:1:1; 1 mark: two \( CH_3 \) groups stated to be equivalent; 1 mark: correct symmetry-based explanation of the equivalence. [4]

In propan-1-ol, \( CH_3CH_2CH_2OH \), all three carbon atoms are chemically distinct (no symmetry relates them), giving 3 \( ^{13}C \) environments/peaks. In propan-2-ol, \( (CH_3)_2CHOH \), a plane of symmetry makes the two terminal \( CH_3 \) carbons equivalent, so only 2 distinct carbon environments are observed (one for the two equivalent \( CH_3 \) carbons, one for the central CH). Since propan-1-ol gives 3 peaks and propan-2-ol gives only 2 peaks, running the \( ^{13}C \) spectrum of an unknown sample and counting the peaks distinguishes the two isomers.

1 mark: propan-1-ol correctly identified as giving 3 peaks; 1 mark: propan-2-ol correctly identified as giving 2 peaks; 1 mark: correct symmetry-based reasoning for the equivalence in propan-2-ol; 1 mark: clear comparative conclusion that peak count distinguishes the isomers. [4]

The molecular formula \( C_4H_8O_2 \) with an ester-region \( CH_2 \) signal near \( \delta 4.1 \) suggests an ester. The singlet at \( \delta 2.05 \) (3H) has no neighbouring protons (n+1=1), consistent with the \( CH_3 \) of an ethanoyl (\( CH_3CO- \)) group, which is isolated from other protons by the carbonyl carbon. The quartet at \( \delta 4.12 \) (2H) is split by 3 equivalent neighbouring protons (n+1=4) and its downfield shift is consistent with \( -O{-}CH_2{-} \) next to the ester oxygen. The triplet at \( \delta 1.25 \) (3H) is split by 2 neighbouring protons (n+1=3), consistent with a \( CH_3 \) adjacent to the \( OCH_2 \) group. Putting these together: \( CH_3COOCH_2CH_3 \), ethyl ethanoate, where the \( OCH_2 \) and terminal \( CH_3 \) are mutually coupled (quartet/triplet pair) and the acetyl \( CH_3 \) is an isolated singlet.

1 mark: correct overall structure, ethyl ethanoate; 1 mark: singlet at δ2.05 correctly assigned to the isolated \( CH_3CO- \) group; 1 mark: quartet at δ4.12 correctly assigned to \( OCH_2 \), split by the adjacent \( CH_3 \); 1 mark: triplet at δ1.25 correctly assigned to \( CH_3 \), split by the adjacent \( CH_2 \); 1 mark: correct name, ethyl ethanoate. [5]

Total moles \( HCl = 0.100 \times 0.200 = 0.0200 \text{ mol} \). Moles \( NaOH = 0.04200 \times 0.100 = 4.20\times10^{-3} \text{ mol} = \) moles excess HCl. Moles \( HCl \) that reacted with \( CaCO_3 = 0.0200 - 4.20\times10^{-3} = 0.0158 \text{ mol} \). Since \( CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \), moles \( CaCO_3 = 0.0158/2 = 7.90\times10^{-3} \text{ mol} \). Mass \( CaCO_3 = 7.90\times10^{-3} \times 100.1 = 0.791 \text{ g} \). Percentage purity \( = \dfrac{0.791}{1.00}\times100 = 79.1\% \).

1 mark: total moles HCl added; 1 mark: moles NaOH = moles excess HCl; 1 mark: moles HCl reacted with \( CaCO_3 \) found by subtraction; 1 mark: correct moles/mass \( CaCO_3 \) (using the 1:2 stoichiometry); 1 mark: correct final percentage 79.1% (accept 78–80%). [5]

NaOH pellets are hygroscopic and readily absorb both water vapour and \( CO_2 \) from the air, so the mass weighed out is not accurately known to be pure, anhydrous NaOH; the exact concentration of the solution therefore cannot be relied upon from the mass alone and must instead be found by titration (standardisation) against a primary standard. A suitable primary standard is potassium hydrogenphthalate (KHP, \( C_8H_5KO_4 \)) or anhydrous sodium carbonate. A primary standard must be obtainable in a very high state of purity, and must be stable — non-hygroscopic and not decomposing or reacting with substances in the air — so that an accurately weighed mass corresponds exactly to a known number of moles.

1 mark: correct reason NaOH is unsuitable (hygroscopic/absorbs CO₂, mass not accurately pure NaOH); 1 mark: valid named primary standard (e.g. KHP or anhydrous Na₂CO₃); 1 mark: property — high purity; 1 mark: property — stable/non-hygroscopic. [4]

\( C_1V_1 = C_2V_2 \): \( V_1 = \dfrac{0.0500\times250}{0.200} = 62.5 \text{ cm}^3 \). Using a graduated pipette (and pipette filler), 62.5 cm³ of the stock \( KMnO_4 \) solution is measured accurately and transferred into a clean 250 cm³ volumetric flask. Distilled water is added with swirling to dissolve/mix, up to (but not exceeding) the graduation mark, using a dropping pipette for the final few drops to accurately reach the bottom of the meniscus on the mark; the flask is then stoppered and inverted several times to ensure the solution is thoroughly and uniformly mixed.

1 mark: correct volume of stock solution, 62.5 cm³; 1 mark: pipette (with filler) correctly named for measuring the stock solution; 1 mark: 250 cm³ volumetric flask correctly named and made up to the graduation mark; 1 mark: correct detail on accurate meniscus reading and thorough mixing by inversion. [4]

Two burette readings are needed (initial and final), each with an uncertainty of ±0.05 cm³, giving a total absolute uncertainty of ±0.10 cm³. Percentage uncertainty \( = \dfrac{0.10}{24.60}\times100 = 0.41\% \). This could be reduced by (i) using a larger titre volume — for example by using a lower concentration of standard solution or a larger/less concentrated sample so that a greater volume of titrant is required — so that the same fixed absolute uncertainty represents a smaller percentage of a larger titre; and (ii) repeating the titration and calculating a mean from concordant results (titres agreeing within 0.10 cm³), which reduces the effect of random reading error on the final result.

1 mark: correct total absolute uncertainty (±0.10 cm³); 1 mark: correct percentage uncertainty (0.41%); 1 mark: valid method 1 to reduce uncertainty (larger titre volume); 1 mark: valid method 2 to reduce uncertainty (repeat and average concordant titres). [4]

A mole fraction of \( Fe^{3+} \) equal to 0.50 means the mixture giving maximum absorbance contains equal moles of \( Fe^{3+} \) and \( SCN^- \), so the ratio of metal to ligand in the complex is 1:1; the formula is \( [Fe(SCN)]^{2+} \). Absorbance (proportional to the concentration of the coloured complex) is greatest at this mole fraction because this is the exact ratio in which the two reagents are combined in the same proportion as in the complex itself, allowing the maximum possible concentration of complex to form with neither reagent left significantly in excess; at any other mole fraction, one reagent is in excess and correspondingly less complex (and hence less absorbance) is formed.

1 mark: 1:1 ratio correctly deduced from mole fraction 0.50; 1 mark: correct formula \( [Fe(SCN)]^{2+} \); 1 mark: explanation that maximum mole fraction corresponds to the same ratio as in the complex; 1 mark: explanation that excess reagent at other ratios reduces the amount of complex formed. [4]

\( c(\text{diluted}) = 0.646/15.2 = 0.0425 \text{ mol dm}^{-3} \). Using \( C_1V_1 = C_2V_2 \): \( C_1 = \dfrac{0.0425 \times 100}{10.0} = 0.425 \text{ mol dm}^{-3} \). Dilution was necessary because the original, more concentrated solution would give an absorbance reading outside the linear (Beer–Lambert law) range of the calibration curve; diluting the sample brings its absorbance down within the calibrated linear range so that the calibration equation can be applied accurately.

1 mark: correct concentration of the diluted solution; 1 mark: correct dilution calculation using \( C_1V_1=C_2V_2 \); 1 mark: correct final concentration 0.425 mol dm⁻³; 1 mark: correct reason for dilution (staying within the linear/Beer–Lambert range of the calibration curve). [4]

\( 2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2 \); \( 2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-} \). This mechanism increases the rate because \( Fe^{3+}/Fe^{2+} \) can react with each anion separately, via one-electron-transfer steps, rather than requiring the two negatively charged ions \( S_2O_8^{2-} \) and \( I^- \) to collide directly with one another; avoiding a direct anion–anion collision removes the very high activation energy associated with electrostatic repulsion between two negative ions, so the catalysed pathway has a much lower activation energy and proceeds faster than the uncatalysed reaction. The iron species is regenerated as \( Fe^{3+} \) at the end of the two-step cycle, confirming its role as a true catalyst. This is an example of homogeneous catalysis because the catalyst (\( Fe^{3+}/Fe^{2+} \)) and the reactants are all present in the same phase (aqueous solution).

1 mark: correct first half-equation; 1 mark: correct second half-equation, regenerating \( Fe^{3+} \); 1 mark: explanation that the mechanism avoids direct anion–anion collision/repulsion; 1 mark: explanation that this lowers the activation energy, increasing rate; 1 mark: correct definition of homogeneous catalysis (catalyst and reactants in the same phase). [5]

In an octahedral complex, the approach of six ligands causes the five degenerate 3d orbitals of the metal ion to split into two energy levels — a lower-energy set of three orbitals and a higher-energy set of two — separated by an energy gap \( \Delta E \). If the d subshell is partially filled, as in \( Cu^{2+} \) (\( 3d^9 \)), an electron can absorb a photon of visible light with energy exactly equal to \( \Delta E \) and be promoted from the lower to the higher set of d orbitals (a d–d transition); the wavelengths of visible light that are not absorbed are transmitted, giving the complex its observed colour (the complementary colour to that absorbed). \( Sc^{3+} \) has an empty \( 3d^0 \) configuration, so there are no d electrons available to undergo a d–d transition between the split orbitals; consequently no visible light is absorbed by this mechanism, and \( Sc^{3+} \) complexes are colourless.

1 mark: octahedral ligand field splits the 3d orbitals into two sets separated by ΔE; 1 mark: partially filled d subshell allows an electron to be promoted (d–d transition) by absorbing visible light of energy ΔE; 1 mark: colour observed is due to the wavelengths not absorbed (complementary colour); 1 mark: \( Sc^{3+} \) has a \( 3d^0 \) configuration, no d electrons to promote; 1 mark: correct overall conclusion that no d–d transition is possible so no visible light is absorbed. [5]

\( E^{\ominus}_{cell} = E^{\ominus}(\text{reduction}) - E^{\ominus}(\text{oxidation}) = 0.77 - 0.54 = +0.23 \text{ V} \); since \( E^{\ominus}_{cell} > 0 \), the reaction is thermodynamically feasible, so \( Fe^{3+} \) will oxidise \( I^- \) to \( I_2 \). Half-equations: \( Fe^{3+} + e^- \rightarrow Fe^{2+} \) (reduction, occurs as written); \( 2I^- \rightarrow I_2 + 2e^- \) (oxidation). Combining (doubling the iron half-equation to balance electrons): \( 2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2 \).

1 mark: correct \( E^{\ominus}_{cell} = +0.23 \text{ V} \) with correct sign; 1 mark: correct feasibility conclusion (yes, Fe³⁺ oxidises I⁻); 1 mark: two correct half-equations; 1 mark: correctly balanced overall ionic equation. [4]

(1) Kinetic barrier: the reaction may have a high activation energy, so even though it is thermodynamically favourable it proceeds extremely slowly (a kinetic, rather than thermodynamic, limitation). (2) Non-standard conditions: standard electrode potentials apply only under standard conditions (298 K, 1 mol dm⁻³, 100 kPa); if the actual concentrations, pressures or temperature differ substantially from standard, the real electrode potentials — and hence the feasibility or direction of the reaction — may differ from the prediction. (3) Competing/side reactions: an alternative reaction with a more favourable (more positive) cell potential may occur preferentially, consuming the reactants before the predicted reaction can proceed to a significant extent.

1 mark: kinetic/activation energy barrier correctly explained; 1 mark: non-standard conditions correctly explained; 1 mark: competing/side reactions correctly explained; 1 mark: overall clear, chemically accurate use of terminology linking at least one reason explicitly to the discrepancy between prediction and observation. [4]

\( CH_3CH_2Br + 2NH_3 \rightarrow CH_3CH_2NH_2 + NH_4Br \) (the HBr initially formed is neutralised by excess ammonia). The mechanism is nucleophilic substitution ( \(S_N2\) ), with the lone pair on the ammonia nitrogen attacking the electrophilic carbon bonded to the halogen. A large excess of ammonia is used because the primary amine product, ethylamine, is itself a nucleophile and can react further with unreacted bromoethane to form secondary and tertiary amines and ultimately a quaternary ammonium salt; using a large excess of ammonia relative to bromoethane makes it statistically far more likely that bromoethane molecules will collide with (unreacted) ammonia rather than with the amine product, favouring mono-substitution and maximising the yield of the primary amine.

1 mark: correct overall equation; 1 mark: mechanism correctly named as nucleophilic substitution; 1 mark: reasoning that the amine product can itself act as a nucleophile leading to further substitution; 1 mark: correct explanation that excess ammonia favours mono-substitution/primary amine. [4]

(a) \( C_6H_5NH_2 + HNO_2 + HCl \rightarrow C_6H_5N_2^+Cl^- + 2H_2O \). (b) An azo compound (azo dye) is formed, joined by an \( -N=N- \) azo linkage via electrophilic substitution of the diazonium ion onto the activated phenol ring; such azo dyes are typically strongly coloured, often orange or yellow-orange. (c) The temperature must be kept below 5 °C because the diazonium salt is thermally unstable and decomposes above about 5–10 °C, releasing nitrogen gas and forming a phenol (via reaction with water) instead of remaining as the diazonium ion needed for the coupling reaction.

1 mark: correct equation for diazonium salt formation; 1 mark: azo compound/dye correctly named as the coupling product with the −N=N− linkage; 1 mark: correct typical colour (orange/yellow); 1 mark: correct explanation that low temperature prevents thermal decomposition of the diazonium salt. [4]

A few drops of ethylamine solution are added to red litmus paper (or to universal indicator solution); the litmus turns blue (or the indicator turns purple/blue, pH > 7), showing the solution is basic/alkaline: \( CH_3CH_2NH_2 + H_2O \rightleftharpoons CH_3CH_2NH_3^+ + OH^- \). The ethyl group is electron-donating by induction (the \( +I \) effect), pushing electron density onto the nitrogen atom; this makes the nitrogen's lone pair more available/more negatively polarised and therefore better able to accept a proton than the lone pair on ammonia's nitrogen, so a greater proportion of ethylamine molecules are protonated at a given pH than ammonia molecules, making ethylamine the stronger base (larger \( K_b \)).

1 mark: correct reagent and observation (litmus/indicator turns blue/basic colour); 1 mark: correct equilibrium equation showing base behaviour; 1 mark: correct +I/electron-donating reasoning for the ethyl group; 1 mark: correct comparative conclusion (ethylamine stronger base, larger Kb, than ammonia). [4]

\( CH_3CH_2CN + 4[H] \rightarrow CH_3CH_2CH_2NH_2 \). Strictly anhydrous/dry conditions are required because \( LiAlH_4 \) reacts violently (often explosively) with water and other protic solvents, releasing flammable hydrogen gas and being destroyed before it can reduce the nitrile. Advantage: converting a halogenoalkane to a nitrile (using \( KCN \) in ethanol) and then reducing the nitrile extends the carbon chain by one carbon atom; this gives a primary amine containing one more carbon atom than the starting halogenoalkane, which cannot be achieved by direct nucleophilic substitution of the halogenoalkane with ammonia (which keeps the same number of carbon atoms).

1 mark: correct equation using [H] notation; 1 mark: correct reason for anhydrous conditions (violent reaction with water/protic solvents); 1 mark: correct identification that the nitrile route extends the carbon chain by one carbon; 1 mark: correct comparison with direct ammonia substitution (no chain extension). [4]

(i) \( CH_3CONH_2 + HCl + H_2O \rightarrow CH_3COOH + NH_4Cl \); the nitrogen-containing product is the ammonium ion, \( NH_4^+ \) (present as ammonium chloride in solution), because the ammonia formed by hydrolysis is immediately protonated by the excess strong acid present. (ii) \( CH_3CONH_2 + NaOH \rightarrow CH_3COONa + NH_3 \); the nitrogen-containing product is ammonia gas, \( NH_3 \), which is released on warming (turning damp red litmus paper blue).

1 mark: correct equation for acid hydrolysis; 1 mark: correct product identified as \( NH_4^+ \) with reasoning (protonation by excess acid); 1 mark: correct equation for alkaline hydrolysis; 1 mark: correct product identified as \( NH_3 \) gas. [4]

\( pI = \dfrac{pK_{a1}+pK_{a2}}{2} = \dfrac{2.34+9.60}{2} = 5.97 \). The pI is the pH at which the amino acid carries zero overall net charge, existing predominantly as the zwitterion. This occurs midway between the two \( pK_a \) values that flank the zwitterion form on a titration curve — the \( pK_a \) of the \( -COOH \) group, below which the fully protonated cationic form predominates, and the \( pK_a \) of the \( -NH_3^+ \) group, above which the fully deprotonated anionic form predominates — because it is at this midpoint pH that the concentrations of the cationic and anionic forms are equal, so their charges exactly cancel, leaving the zwitterion as the dominant species with net zero charge.

1 mark: correct formula, pI = average of the two pKa values; 1 mark: correct substitution and final answer 5.97; 1 mark: pI correctly linked to the zwitterion/zero net charge; 1 mark: correct reasoning that at the midpoint the cationic and anionic forms are present in equal, charge-cancelling amounts. [4]

Alanine has pI = 6.0, exactly equal to the buffer pH, so it carries zero net charge at this pH and does not migrate towards either electrode. Lysine has pI = 9.7, which is above the buffer pH of 6.0; since the pH is below its pI, lysine is predominantly protonated and carries an overall positive charge, so it migrates towards the negative electrode (cathode). Aspartic acid has pI = 2.8, which is below the buffer pH of 6.0; since the pH is above its pI, aspartic acid is predominantly deprotonated and carries an overall negative charge, so it migrates towards the positive electrode (anode).

1 mark: alanine correctly shown not to migrate, with reasoning (pH = pI, zero net charge); 1 mark: lysine correctly shown to migrate to the cathode; 1 mark: aspartic acid correctly shown to migrate to the anode; 1 mark: correct general reasoning throughout comparing buffer pH with each amino acid's pI to determine charge. [4]

\( H_2NCH_2COOH + H_2NCH(CH_3)COOH \rightarrow H_2NCH_2CONHCH(CH_3)COOH + H_2O \). This is a condensation reaction, since a small molecule (water) is lost as the two amino acids join through a new amide (peptide) bond. From a mixture of glycine (Gly) and alanine (Ala), 4 different dipeptides can form: Gly-Gly, Ala-Ala, Gly-Ala and Ala-Gly, since the order in which the two amino acids are joined affects which end of the dipeptide carries the free \( -NH_2 \) group and which carries the free \( -COOH \) group.

1 mark: correct equation showing formation of the peptide bond and loss of water; 1 mark: reaction correctly named as condensation; 1 mark: correct number of dipeptides, 4; 1 mark: correct reasoning that order/direction of joining gives distinct dipeptides (Gly-Ala ≠ Ala-Gly). [4]

(a) Repeat unit: \( -[O{-}CH(CH_3){-}CO]- \), formed as an ester linkage between the \( -OH \) and \( -COOH \) groups of successive lactic acid monomers, with loss of water at each linkage. (b) PLA contains polar ester linkages throughout its backbone; these can be hydrolysed by water and broken down further by microbial/enzymatic action in the environment, degrading the polymer back into smaller molecules (ultimately lactic acid, then \( CO_2 \) and \( H_2O \)). Poly(ethene), by contrast, has a saturated hydrocarbon backbone made entirely of non-polar \( C{-}C \) and \( C{-}H \) bonds, with no hydrolysable linkages; it is therefore chemically inert to water and resistant to microbial attack, making it non-biodegradable.

1 mark: correct repeat unit of PLA; 1 mark: ester linkage in PLA correctly identified as hydrolysable/polar; 1 mark: correct explanation of microbial/hydrolytic breakdown of PLA; 1 mark: correct contrast with poly(ethene)'s non-polar, non-hydrolysable C−C backbone. [4]

A known volume (e.g. 25.0 cm³, measured using a pipette) of the tap water sample is placed in a conical flask, and a few cm³ of ammonia/ammonium chloride buffer solution is added to maintain the pH at approximately 10, which is necessary both for the indicator to respond correctly and for complexation with EDTA to be favourable. A small amount of solid Eriochrome Black T indicator is then added, which forms a wine-red/pink complex with any free \( Ca^{2+} \) and \( Mg^{2+} \) ions present. Standardised EDTA solution is added from a burette to the flask with continuous swirling; EDTA forms a much more stable 1:1 complex with the metal ions than the indicator does, so as EDTA is added it progressively displaces the indicator from the metal ions. The end point is reached when the last free metal–indicator complex has been displaced, at which point the colour changes sharply from wine-red to blue, showing that all the \( Ca^{2+} \) and \( Mg^{2+} \) ions are now bound to EDTA. The titration is repeated until concordant titres (within 0.10 cm³) are obtained and a mean titre is calculated. Since EDTA reacts with both \( Ca^{2+} \) and \( Mg^{2+} \) in 1:1 stoichiometry, the moles of EDTA used equal the total moles of \( Ca^{2+}+Mg^{2+} \) in the sample, from which the total hardness (often expressed as an equivalent mass concentration of \( CaCO_3 \)) can be calculated.

Band A (5–6 marks): full, coherent account covering the role of the pH 10 buffer, the Eriochrome Black T mechanism (wine-red complex displaced by EDTA), the sharp red-to-blue colour change at the end point, repeat titrations for concordance, and the 1:1 EDTA:metal stoichiometry used to calculate total hardness; fluent use of specialist terms, essentially free from errors in spelling, punctuation and grammar. Band B (3–4 marks): most elements present but less complete or less clearly linked. Band C (1–2 marks): only a fragment described, e.g. 'add EDTA until the colour changes'. Indicative content: buffer at pH ≈ 10; Eriochrome Black T; wine-red to blue colour change; EDTA displaces indicator from metal ions (greater stability constant); concordant titres; 1:1 EDTA:metal stoichiometry used to calculate total hardness. [6]

Solutions of \( Ni^{2+} \) and the ligand L are prepared at equal (equimolar) concentration. A series of mixtures is then made up, keeping the total volume (and hence the total moles of metal plus ligand) constant across all mixtures, but systematically varying the volume ratio of the two solutions (for example 1:9, 2:8, 3:7, up to 9:1), so that the mole fraction of \( Ni^{2+} \) varies from close to 0 to close to 1. The absorbance of each mixture is measured with a colorimeter set to a wavelength strongly absorbed by the complex (chosen from a preliminary absorption spectrum), with the instrument first zeroed against a blank such as distilled water. A graph of absorbance against the mole fraction of \( Ni^{2+} \) is then plotted; the absorbance rises and then falls, forming two roughly straight-line portions on either side of a peak, which are extrapolated until they intersect. The mole fraction of \( Ni^{2+} \) at which the two extrapolated lines intersect corresponds to the stoichiometric ratio of metal to ligand in the complex — for example, an intersection at a mole fraction of 0.33 indicates a metal:ligand ratio of 1:2 — because it is at this exact mixing ratio that the maximum possible concentration of complex forms, with neither reagent in excess, giving the highest absorbance.

Band A (5–6 marks): full account including equimolar solutions, constant total volume with systematically varying ratio, wavelength selection and blanking of the colorimeter, plotting absorbance against mole fraction, extrapolating the two linear portions to their intersection, and correct interpretation of the mole fraction at the intersection as the stoichiometric ratio; fluent, accurate use of specialist terms. Band B (3–4 marks): most elements present but less complete or less clearly sequenced. Band C (1–2 marks): only a fragment described, e.g. 'measure absorbance of different mixtures'. Indicative content: equimolar Ni²⁺ and L solutions; constant total volume, varying ratio; colorimeter wavelength selection and blank; graph of absorbance vs mole fraction; extrapolate two linear portions to their intersection; mole fraction at the intersection = stoichiometric ratio. [6]

Titres: rough = 24.50 − 0.00 = 24.50 cm³; titration 1 = 23.80 − 0.00 = 23.80 cm³; titration 2 = 23.75 − 0.05 = 23.70 cm³; titration 3 = 23.85 − 0.10 = 23.75 cm³. The rough titration is excluded from the mean. Titrations 1, 2 and 3 (23.80, 23.70, 23.75 cm³) all agree with each other to within 0.10 cm³, so all three are concordant. Mean titre \( = \dfrac{23.80+23.70+23.75}{3} = \dfrac{71.25}{3} = 23.75 \text{ cm}^3 \).

1 mark: all four titre values correctly calculated; 1 mark: rough titration correctly excluded; 1 mark: titrations 1–3 correctly identified as concordant (within 0.10 cm³); 1 mark: correct mean titre 23.75 cm³. [4]

Moles \( NaOH = 0.02140 \times 0.100 = 2.140\times10^{-3} \text{ mol} = \) moles \( CH_3COOH \) in the diluted 25.0 cm³ sample (1:1 reaction). Concentration in the diluted sample \( = \dfrac{2.140\times10^{-3}}{0.0250} = 0.0856 \text{ mol dm}^{-3} \). Since the original vinegar was diluted by a factor of 10, its concentration \( = 0.0856 \times 10 = 0.856 \text{ mol dm}^{-3} \). Mass concentration \( = 0.856 \times 60.0 = 51.4 \text{ g dm}^{-3} \).

1 mark: correct moles NaOH = moles ethanoic acid in diluted sample; 1 mark: correct concentration of diluted sample; 1 mark: correct ×10 dilution factor applied to find original concentration; 1 mark: correct final answer 51.4 g dm⁻³ (accept 51–52). [4]

Moles \( KMnO_4 = 0.00910 \times 0.0100 = 9.10\times10^{-5} \text{ mol} \). Moles \( Fe^{2+} \) in the 25.0 cm³ portion \( = 5 \times 9.10\times10^{-5} = 4.55\times10^{-4} \text{ mol} \). Total moles \( Fe^{2+} \) in 250 cm³ \( = 4.55\times10^{-4} \times \dfrac{250}{25.0} = 4.55\times10^{-3} \text{ mol} \). Mass \( Fe = 4.55\times10^{-3} \times 55.8 = 0.254 \text{ g} \). Percentage \( Fe = \dfrac{0.254}{0.500}\times100 = 50.8\% \).

1 mark: correct moles \( KMnO_4 \); 1 mark: correct moles \( Fe^{2+} \) in the 25.0 cm³ portion (×5 stoichiometric factor); 1 mark: correct scale-up (×10) to total moles \( Fe^{2+} \) in 250 cm³; 1 mark: correct final percentage 50.8% (accept 50–52%). [4]

Adding \( NaOH(aq) \) dropwise then in excess to each solution: \( Fe^{2+} \) gives a green (pale green/grey-green) gelatinous precipitate of \( Fe(OH)_2 \), insoluble in excess NaOH, which darkens/turns rust-brown at its surface on standing in air as it is oxidised to \( Fe(OH)_3 \). \( Fe^{3+} \) gives an orange-brown precipitate of \( Fe(OH)_3 \) immediately, insoluble in excess. \( Cu^{2+} \) gives a pale blue gelatinous precipitate of \( Cu(OH)_2 \), insoluble in excess. \( Co^{2+} \) gives a blue precipitate of \( Co(OH)_2 \), insoluble in excess, which may darken/turn pink-brown on standing in air due to slow oxidation.

1 mark: correct observation for \( Fe^{2+} \) (green ppt, darkens in air); 1 mark: correct observation for \( Fe^{3+} \) (orange-brown ppt); 1 mark: correct observation for \( Cu^{2+} \) (pale blue ppt); 1 mark: correct observation for \( Co^{2+} \) (blue ppt, darkens/changes on standing). [4]

On adding aqueous ammonia dropwise, a pale blue gelatinous precipitate of \( Cu(OH)_2 \) forms, as the ammonia acts as a weak base, providing \( OH^- \) ions. On adding a large excess of ammonia, this precipitate dissolves to give a deep (royal) blue solution: the hydroxide precipitate reacts with excess ammonia by ligand substitution, in which four of the six water ligands around the copper ion are replaced by four ammonia ligands, forming the soluble complex ion \( [Cu(NH_3)_4(H_2O)_2]^{2+} \).

1 mark: correct observation on dropwise addition (pale blue precipitate); 1 mark: correct observation in excess (dissolves to deep/royal blue solution); 1 mark: correct explanation as ligand substitution; 1 mark: correct formula of the resulting complex \( [Cu(NH_3)_4(H_2O)_2]^{2+} \). [4]

The sample is treated with sodium nitrite and dilute hydrochloric acid at a temperature kept below 5 °C (in an ice bath), forming a diazonium salt if a primary aromatic amine is present. This solution is then added to an alkaline solution of phenol (or 2-naphthol). If a primary aromatic amine was present, an intensely coloured (orange or yellow, or red with 2-naphthol) azo dye precipitate/solution forms immediately, confirming its presence; a primary aliphatic amine under the same conditions instead forms an unstable diazonium salt that decomposes rapidly (evolving nitrogen gas), giving no coloured azo product.

1 mark: correct reagents and low-temperature condition (NaNO₂/dilute HCl, below 5 °C); 1 mark: correct coupling step and observation (coloured azo dye forms with alkaline phenol); 1 mark: correct distinguishing note that an aliphatic amine gives no coloured product under the same conditions. [3]

The dried chromatogram is sprayed with a solution of ninhydrin, and then gently warmed (for example in an oven, or with a hairdryer) to develop the colour. Spots where amino acids are present on the chromatogram develop a purple (blue-purple) colour, allowing their positions to be located, marked and, together with a solvent front measurement, used to calculate \( R_f \) values for identification by comparison with known reference amino acids.

1 mark: correct reagent, ninhydrin, applied by spraying; 1 mark: correct condition, warming/drying after spraying; 1 mark: correct observation, purple/blue-purple colour develops at amino acid spots. [3]

Cyclohexene has a discrete, localised \( C=C \) double bond with a high, exposed electron density; this induces a temporary dipole in the approaching \( Br_2 \) molecule (making one bromine atom \( \delta^+ \)), which is then attacked directly by the double bond, undergoing rapid electrophilic addition at room temperature without any catalyst, decolourising the bromine and forming 1,2-dibromocyclohexane. In methylbenzene, the ring's \( \pi \) electrons are delocalised over six carbon atoms, giving a much lower electron density at any single point than a localised double bond; \( Br_2 \) alone is not sufficiently polarised to react with the ring at a useful rate, so a halogen-carrier catalyst (e.g. \( AlCl_3 \)) is required to generate a much stronger electrophile. Furthermore, the ring reacts by electrophilic substitution rather than addition, because addition would destroy the delocalised aromatic system and its associated stabilisation (delocalisation energy); substitution allows a hydrogen atom to be replaced while the stable, delocalised ring system is retained.

1 mark: correct mechanism/reasoning for cyclohexene's rapid addition (localised, electron-rich double bond); 1 mark: correct reasoning that the delocalised ring is less reactive, requiring a catalyst to generate a stronger electrophile; 1 mark: correct explanation that substitution (not addition) preserves ring aromaticity/delocalisation energy; 1 mark: overall coherent comparison correctly linking mechanism type to structure in both cases. [4]

(a) The acceleration stage uses an electric field to give all the positively charged ions the same kinetic energy, so that they subsequently travel through the deflection region at different, mass-dependent velocities. (b) The magnetic field exerts a deflecting force on the moving charged ions; for a given (equal) kinetic energy, ions of smaller mass (smaller m/z, since most ions carry a single positive charge) travel faster and are deflected through a larger angle by the magnetic field than heavier ions (larger m/z), which travel more slowly and are deflected less.

1 mark: correct purpose of acceleration (equal kinetic energy for all ions); 1 mark: correct link between acceleration and different velocities depending on mass; 1 mark: correct statement that smaller m/z ions are deflected more; 1 mark: correct reasoning linking mass/velocity to the degree of deflection. [4]

(a) Deuterated trichloromethane (deuterochloroform), \( CDCl_3 \), is commonly used (or deuterated water, \( D_2O \), for water-soluble samples). (b) Tetramethylsilane (TMS), \( Si(CH_3)_4 \), is used as the reference, set at \( \delta = 0 \). It is suitable because: it gives a single, sharp signal (all 12 protons are chemically equivalent); it is chemically inert/unreactive and does not interfere with the sample; it is volatile, so it can be easily removed from the sample after use; and its signal appears well clear of (upfield of) almost all other proton signals, avoiding overlap with the sample's peaks.

1 mark: correct deuterated solvent named (CDCl₃, or D₂O for aqueous samples); 1 mark: TMS correctly named as the reference; 1 mark: valid reason 1 (single sharp signal / all protons equivalent, or chemically inert); 1 mark: valid reason 2 (volatile/easily removed, or signal well separated from other peaks). [4]

(a) Ethenol is the unstable enol tautomer of ethanal; it rapidly and spontaneously isomerises (tautomerises) to the far more stable carbonyl compound, ethanal, \( CH_3CHO \), so it cannot be isolated or stored in sufficient purity/quantity to be polymerised directly. (b) The polyester side-chains are hydrolysed using aqueous sodium hydroxide under reflux (alkaline hydrolysis). For one repeat unit: \( -[CH_2CH(OCOCH_3)]- + NaOH \rightarrow -[CH_2CH(OH)]- + CH_3COONa \).

1 mark: correct explanation that ethenol tautomerises/isomerises to ethanal; 1 mark: correct statement of instability preventing isolation/polymerisation; 1 mark: correct reagent and conditions (NaOH(aq), reflux); 1 mark: correct equation for the repeat-unit conversion, with sodium ethanoate as by-product. [4]

Phenylamine is added to an excess of ethanoic anhydride and the mixture is gently warmed (e.g. on a water bath) for several minutes, allowing the acylation (condensation) reaction to go to completion and form N-phenylethanamide, with ethanoic acid produced as a by-product. The reaction mixture is then poured into a beaker of cold water, which also hydrolyses any unreacted ethanoic anhydride; the crude solid amide, being insoluble in cold water, precipitates out immediately. The crude solid is collected by filtration under reduced pressure, washed with a little cold water to remove soluble impurities, and (if a purer sample is required) purified further by recrystallisation.

1 mark: correct reagents (phenylamine + excess ethanoic anhydride); 1 mark: correct conditions (gentle warming); 1 mark: correct isolation step (pour into cold water, precipitation, also hydrolysing excess anhydride); 1 mark: correct final collection (filtration under reduced pressure, washing, optional recrystallisation). [4]

Mass of methylbenzene \( = 5.00 \times 0.867 = 4.335 \text{ g} \). Moles methylbenzene \( = 4.335/92.0 = 0.04712 \text{ mol} \). Since the oxidation is 1:1, theoretical moles of benzoic acid = 0.04712 mol, giving a theoretical mass \( = 0.04712 \times 122.0 = 5.75 \text{ g} \). Percentage yield \( = \dfrac{4.50}{5.75}\times100 = 78.3\% \).

1 mark: correct mass of methylbenzene from density; 1 mark: correct moles of methylbenzene; 1 mark: correct theoretical mass of benzoic acid (5.75 g); 1 mark: correct final percentage yield 78.3% (accept 77–79%). [4]

The crude mixture is shaken with (or run into) an excess of sodium carbonate solution, which neutralises the sulfuric acid catalyst and any unreacted ethanoic acid (releasing \( CO_2 \)) and dissolves the polar methanol into the aqueous layer; the ester, being only sparingly soluble in water, forms a separate, immiscible organic layer, which is run off using a separating funnel. This organic layer is dried using an anhydrous drying agent such as anhydrous magnesium sulfate, which is then removed by gravity filtration. The pure ester is finally isolated by careful fractional distillation of the dried liquid, collecting the fraction that distils over close to 57 °C.

1 mark: correct washing step to remove acid/catalyst (sodium carbonate solution); 1 mark: correct use of a separating funnel, based on the ester's immiscibility with water; 1 mark: correct drying step (anhydrous MgSO₄, filtered off); 1 mark: correct final fractional distillation collecting the fraction at ~57 °C. [4]

Benzene is added slowly, with continuous shaking or stirring, to a cooled mixture of concentrated nitric acid and concentrated sulfuric acid (the 'nitrating mixture'), with the reaction flask kept in an ice/water bath to control the temperature below about 55 °C. After addition, the mixture may be gently warmed briefly to complete the reaction, then poured into cold water; the dense, oily nitrobenzene layer is separated using a separating funnel, washed with sodium carbonate solution to remove residual acid, dried, and purified. The temperature must be carefully controlled because the nitration reaction is strongly exothermic; at higher temperatures, further substitution can occur, producing unwanted dinitrobenzene as a by-product (poly-nitration), and there is also a greater risk of the strongly exothermic reaction becoming vigorous/uncontrolled ('running away').

1 mark: correct procedure (slow addition of benzene to cooled nitrating mixture, ice bath); 1 mark: correct work-up (separating funnel, washing, drying); 1 mark: correct reason — prevents dinitration/poly-substitution at higher temperature; 1 mark: correct reason — prevents a runaway/uncontrolled exothermic reaction. [4]

Iodine is continuously produced by the reaction \( H_2O_2 + 2I^- + 2H^+ \rightarrow I_2 + 2H_2O \). However, this iodine is immediately and rapidly consumed by reaction with the small, fixed quantity of thiosulfate present: \( I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-} \); as a result, no free iodine — and therefore no blue-black starch–iodine colour — can build up while any thiosulfate remains unreacted. Only once all of the thiosulfate has been consumed does any further iodine produced remain in solution, immediately complexing with the starch indicator to give a sudden, sharp colour change from colourless to blue-black. The time taken for this colour change (the 'clock time') is inversely related to the initial rate of the peroxide–iodide reaction — a shorter clock time indicates a faster initial rate.

1 mark: correct first equation (H₂O₂ + I⁻ producing I₂); 1 mark: correct second equation (I₂ + thiosulfate); 1 mark: correct explanation of why no colour appears until thiosulfate is exhausted; 1 mark: correct link between the clock time and the initial rate of reaction. [4]

The glass electrode is first rinsed with distilled water and gently blotted dry (not rubbed, to avoid damaging the delicate glass bulb). The meter is then calibrated using at least two standard buffer solutions of accurately known pH that bracket the expected range of the samples (for example pH 4 and pH 7, or pH 7 and pH 10): the electrode is placed in the first standard buffer and the meter adjusted to read that exact known pH; the electrode is rinsed again and placed in the second standard buffer, and the meter's slope is adjusted so that it correctly reads this second known pH. The temperature of the buffers and samples should also be noted (or a temperature-compensating probe used), since pH readings are temperature-dependent.

1 mark: correct rinsing/blotting of the electrode between solutions; 1 mark: correct two-point calibration method described; 1 mark: correct use of named standard buffers bracketing the expected range; 1 mark: correct note on temperature dependence/compensation. [4]

Adding NaOH reacts with/removes \( H^+ \) ions, decreasing \( [H^+] \); by Le Chatelier's principle, the equilibrium position shifts to the right (towards the products) to partially replace the \( H^+ \) ions removed, producing more yellow \( CrO_4^{2-}(aq) \) and turning the solution yellow. Adding \( H_2SO_4 \) increases \( [H^+] \); the equilibrium position shifts to the left (towards the reactants) to remove some of the added \( H^+ \) ions (reforming water and dichromate), producing more orange \( Cr_2O_7^{2-}(aq) \) and turning the solution back to orange.

1 mark: correct direction of shift on adding NaOH (right); 1 mark: correct Le Chatelier reasoning and colour for NaOH addition; 1 mark: correct direction of shift on adding H₂SO₄ (left); 1 mark: correct Le Chatelier reasoning and colour for H₂SO₄ addition. [4]

A zinc electrode is placed in \( 1 \text{ mol dm}^{-3} \) \( ZnSO_4(aq) \) at 298 K, and connected via a salt bridge (e.g. filter paper soaked in saturated \( KNO_3(aq) \), or a \( KNO_3 \)-agar gel bridge) to a standard hydrogen electrode (a platinum electrode in \( 1 \text{ mol dm}^{-3} \: H^+(aq) \), with \( H_2 \) gas at 100 kPa bubbled over it, at 298 K). The two electrodes are connected externally through a high-resistance voltmeter. The salt bridge completes the electrical circuit by allowing ions to migrate between the two half-cells to maintain electrical neutrality as the cell reaction proceeds, without allowing the two solutions to mix directly (which would cause unwanted side reactions). A high-resistance voltmeter is used, rather than an ammeter, so that negligible current is drawn from the cell while the potential difference is measured, ensuring the reading obtained is the true equilibrium (reversible) cell potential rather than a lower value caused by current flowing and the cell doing work.

1 mark: correct half-cell set-up (Zn electrode in 1 mol dm⁻³ ZnSO₄, standard hydrogen electrode correctly described); 1 mark: correct role of the salt bridge (completes circuit/maintains neutrality without mixing solutions); 1 mark: correct reason for a high-resistance voltmeter (negligible current drawn); 1 mark: correct link to obtaining the true/reversible standard electrode potential. [4]

Two reasons: (1) heat is lost to the surroundings, by convection, radiation and conduction, from the flame and from the sides of the calorimeter, so not all of the heat released by combustion is transferred to the water being heated; (2) incomplete combustion of the ethanol may occur (producing some carbon monoxide or soot rather than fully oxidising all the carbon to \( CO_2 \)), which releases less energy than complete combustion, and some fuel may also be lost by evaporation before or during burning. Modification: use a bomb calorimeter (a sealed, insulated, oxygen-rich container) in place of simple open calorimetry, which minimises heat loss to the surroundings and ensures more complete combustion, giving a value much closer to the true enthalpy of combustion.

1 mark: valid reason 1 (heat lost to surroundings by convection/radiation/conduction); 1 mark: valid reason 2 (incomplete combustion or fuel evaporation); 1 mark: valid modification suggested (e.g. bomb calorimeter, or insulation/draught shield with a lid); 1 mark: correct reasoning linking the modification to reduced heat loss/more complete combustion. [4]

A back-titration is used when a direct titration is not possible or not accurate — for example when the substance being analysed is insoluble in water (such as calcium carbonate), reacts too slowly with a titrant for a sharp end point to be seen, or when no suitable indicator exists for the direct reaction. In such cases, a known, deliberately excess amount of a standard reactant (e.g. a standard acid) is added, which completely reacts with all of the substance being analysed; the amount of this reactant remaining unreacted (in excess) is then determined by titrating it against a second standard solution (e.g. a standard base). By calculating the total amount of the first reactant added and subtracting the amount found to be in excess (from the second titration), the amount that reacted with the original substance — and hence its concentration or purity — can be calculated.

1 mark: correct reason a back-titration is needed (insoluble/slow-reacting/no suitable indicator); 1 mark: correct example given (e.g. CaCO₃ purity); 1 mark: correct description of adding known excess reagent, then titrating the remainder; 1 mark: correct explanation of the subtraction logic (total added − excess = amount reacted). [4]

Moles \( H_2SO_4 = 0.0500 \times 1.00 = 0.0500 \text{ mol} \); since \( CuO \) is in excess, \( H_2SO_4 \) is the limiting reagent. \( CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O \) is 1:1, so moles \( CuSO_4\cdot5H_2O \) crystallised = 0.0500 mol. Maximum mass \( = 0.0500 \times 249.7 = 12.5 \text{ g} \). The filtered blue solution is gently heated (e.g. in an evaporating basin) to evaporate off some water, concentrating it to the point of saturation (tested by dipping a glass rod in the solution and checking that crystals form on cooling, or until a crystalline film appears at the surface). The hot saturated solution is then left to cool slowly and undisturbed to room temperature, allowing crystals to grow; the crystals are collected by filtration, washed with a small volume of ice-cold water, and dried between sheets of filter paper (not in a hot oven, which would drive off the water of crystallisation).

1 mark: correct identification of H₂SO₄ as the limiting reagent, with correct moles; 1 mark: correct theoretical mass of crystals, 12.5 g; 1 mark: correct evaporation-to-saturation procedure; 1 mark: correct slow cooling/filtration/gentle drying procedure (avoiding loss of water of crystallisation). [4]

Phenylamine is immiscible (or only sparingly miscible) with water. In steam distillation, because the two liquids are immiscible, each exerts its own vapour pressure independently of the other; the mixture boils, and distils over, once the sum of the two vapour pressures equals atmospheric pressure. This occurs at a temperature below 100 °C — well below phenylamine's normal boiling point of 184 °C — allowing phenylamine to be distilled over gently at this lower temperature, without exposing it to the much higher temperature at which it might otherwise decompose, oxidise or discolour. Steam distillation also cleanly separates the volatile phenylamine from the non-volatile inorganic tin salts left behind in the reaction flask.

1 mark: correct statement that phenylamine and water are immiscible; 1 mark: correct explanation that immiscible liquids exert independent vapour pressures, allowing boiling below 100 °C; 1 mark: correct link to avoiding decomposition at 184 °C; 1 mark: correct point on clean separation from non-volatile tin salts. [4]