2022 HKDSE Chemistry Answers & Marking Scheme

2-hydroxybutanoic acid (\(\text{CH}_3\text{CH}_2\text{CH(OH)COOH}\)) is a bifunctional monomer containing both a hydroxyl group and a carboxyl group. Condensation polymerisation occurs with the elimination of water molecules to form a polyester. Thus, (1) is correct because it contains ester linkages. (2) is correct because polyesters with aliphatic backbones are susceptible to hydrolysis of ester groups by water or enzymes, making them biodegradable. (3) is incorrect because condensation polymerisation involves the loss of \(\text{H}_2\text{O}\), so the empirical formula of the polymer repeating unit (\(\text{C}_4\text{H}_6\text{O}_2\)) is different from that of the monomer (\(\text{C}_4\text{H}_8\text{O}_3\)).

Award 1 mark for the correct option A.

The chemical equation for the standard enthalpy of formation of propan-2-ol is: \(3\text{C(graphite)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{O(l)}\). Using Hess's law: \(\Delta H_f^\theta [\text{C}_3\text{H}_8\text{O(l)}] = 3 \times \Delta H_c^\theta [\text{C}] + 4 \times \Delta H_c^\theta [\text{H}_2] - \Delta H_c^\theta [\text{C}_3\text{H}_8\text{O(l)}]\). Substituting the values: \(\Delta H_f^\theta = 3(-394) + 4(-286) - (-2006) = -1182 - 1144 + 2006 = -320\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option A.

To prepare ethyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3\)) from propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)): First, propan-1-ol is oxidized to propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) by heating under reflux with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\text{(aq)}\). Second, the isolated propanoic acid is reacted with ethanol in the presence of concentrated sulfuric acid (catalyst and dehydrating agent) under reflux to form ethyl propanoate via esterification. Thus, option A is correct.

Award 1 mark for the correct option A.

\(\text{BF}_3\) has a trigonal planar shape with a bond angle of \(120^\circ\). \(\text{CH}_4\) has a tetrahedral shape with a bond angle of \(109.5^\circ\). \(\text{NH}_3\) has a trigonal pyramidal shape with a bond angle of \(107^\circ\) due to lone pair-bonding pair repulsion being greater than bonding pair-bonding pair repulsion. \(\text{H}_2\)O has a bent shape with a bond angle of \(104.5^\circ\) due to two lone pairs. Therefore, the decreasing order of bond angle is \(\text{BF}_3 > \text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}\).

Award 1 mark for the correct option A.

During the electrolysis of concentrated \(\text{NaCl(aq)}\): At the cathode (negative electrode), \(\text{H}^+\text{(aq)}\) ions from water are preferentially discharged over \(\text{Na}^+\text{(aq)}\) because hydrogen is lower in the electrochemical series. The reduction of water (\(2\text{H}_2\text{O(l)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)}\)) consumes \(\text{H}^+\) and leaves an excess of \(\text{OH}^-\), which increases the pH around the cathode. Thus, C is correct. At the anode (positive electrode), \(\text{Cl}^-\) is preferentially discharged over \(\text{OH}^-\) because of its high concentration, yielding chlorine gas (not oxygen). Thus, A, B, and D are incorrect.

Award 1 mark for the correct option C.

According to the Arrhenius equation: \(\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A\). Thus, the slope of the plot of \(\ln k\) against \(\frac{1}{T}\) is equal to \(-\frac{E_a}{R}\). Given \(\text{slope} = -1.20 \times 10^4\text{ K}\): \(-\frac{E_a}{R} = -1.20 \times 10^4\text{ K}\) which leads to \(E_a = 1.20 \times 10^4\text{ K} \times 8.31\text{ J mol}^{-1}\text{ K}^{-1} = 99720\text{ J mol}^{-1} \approx 99.7\text{ kJ mol}^{-1}\). Option D has the correct numerical value but the wrong units (\(\text{kJ mol}^{-1}\) instead of \(\text{J mol}^{-1}\)). Thus, option B is correct.

Award 1 mark for the correct option B.

(1) is correct: \(\text{Cu}^{2+}\) has a \(3\text{d}^9\) configuration. In aqueous solution, d-orbitals split, and absorbing orange-red light excites an electron from a lower to a higher d-orbital (d-d transition), leaving the complementary blue colour visible. (2) is incorrect: A transition metal is defined as an element that forms at least one stable ion with an incomplete d subshell. Zinc only forms \(\text{Zn}^{2+}\) with a completely filled d subshell (\(3\text{d}^{10}\)), so it is not a transition metal. (3) is correct: The 4s and 3d subshells are very close in energy, allowing variable numbers of electrons to be lost from both subshells to yield different oxidation states. Thus, (1) and (3) only are correct.

Award 1 mark for the correct option B.

All four compounds have similar relative molecular masses (\(M_r \approx 58 - 60\)), so their dispersion forces (Van der Waals' forces) are of similar strength. Propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) has a highly polar \(\text{O-H}\) bond which allows molecules to form strong intermolecular hydrogen bonds. Methoxyethane and propanal are polar but cannot form hydrogen bonds with themselves (only weaker dipole-dipole forces exist). Butane is non-polar and only experiences weak dispersion forces. Thus, propan-1-ol has the highest boiling point.

Award 1 mark for the correct option A.

In an alkaline hydrogen-oxygen fuel cell: At the anode (negative electrode), hydrogen gas is oxidized: \(\text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + 2\text{e}^-\). At the cathode (positive electrode), oxygen gas is reduced: \(\text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^- \rightarrow 4\text{OH}^-\text{(aq)}\). Thus, option B is correct and A is incorrect. Electrons flow from the anode to the cathode through the external circuit, so C is incorrect. The overall reaction is \(2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}\). Hydroxide ions are consumed at the anode and regenerated at the cathode at the same rate, so the pH of the KOH electrolyte remains strongly alkaline and does not decrease significantly, making D incorrect.

Award 1 mark for the correct option B.

1. Moles of initial \(\text{HCl} = 0.500 \times 0.0500 = 0.0250\text{ mol}\). 2. Moles of excess \(\text{HCl} = \text{moles of NaOH} = 0.250 \times 0.0400 = 0.0100\text{ mol}\). 3. Moles of \(\text{HCl}\) reacted with \(\text{CaCO}_3 = 0.0250 - 0.0100 = 0.0150\text{ mol}\). 4. From equation \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\), moles of \(\text{CaCO}_3 = \frac{1}{2} \times 0.0150 = 0.00750\text{ mol}\). 5. Mass of \(\text{CaCO}_3 = 0.00750 \times 100.1 = 0.75075\text{ g}\). 6. Percentage by mass = \(\frac{0.75075\text{ g}}{1.00\text{ g}} \times 100\% = 75.1\%\). Thus, option C is correct.

Award 1 mark for the correct option C.

The chemical equation for the standard enthalpy change of formation of propene is: \(3\text{C}(s) + 3\text{H}_2(g) \rightarrow \text{C}_3\text{H}_6(g)\). According to Hess's Law: \(\Delta H_f^\ominus[\text{C}_3\text{H}_6(g)] = 3 \times \Delta H_c^\ominus[\text{C}(s)] + 3 \times \Delta H_c^\ominus[\text{H}_2(g)] - \Delta H_c^\ominus[\text{C}_3\text{H}_6(g)] = 3(-394) + 3(-286) - (-2058) = -1182 - 858 + 2058 = +18\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option A.

Statement (1) is correct because the polymer contains the amide linkage (\text{—CONH—}). Statement (2) is incorrect because it is a condensation polymer, releasing water during formation. Statement (3) is correct because the monomers required to form this polyamide (Nylon 6,6) are hexanedioic acid (providing the \(\text{—CO—(CH}_2)_4\text{—CO—}\) part) and hexane-1,6-diamine (providing the \(\text{—NH—(CH}_2)_6\text{—NH—}\) part).

Award 1 mark for the correct option B.

Reaction of propene with \text{HBr} yields 2-bromopropane as the major product (Markovnikov's rule), so X is \(\text{CH}_3\text{CH(Br)CH}_3\). Heating propan-2-ol (a secondary alcohol) with acidified potassium dichromate oxidizes it to propanone (Y), which has the formula \(\text{CH}_3\text{COCH}_3\).

Award 1 mark for the correct option B.

\(\text{CO}_2\) is linear (bond angle \(180^\circ\)). \(\text{CH}_4\) is tetrahedral (bond angle \(109.5^\circ\)). \(\text{NH}_3\) is trigonal pyramidal (bond angle \(107^\circ\) due to one lone pair). \(\text{H}_2\text{O}\) is V-shaped (bond angle \(104.5^\circ\) due to two lone pairs). Therefore, the correct order of decreasing bond angle is \(\text{CO}_2 > \text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}\).

Award 1 mark for the correct option A.

Statement (1) is correct because chloride ions are preferentially discharged at the anode to form chlorine gas (a pale green choking gas). Statement (2) is correct because hydrogen ions are discharged at the cathode, leaving hydroxide ions in the solution, which increases the pH. Statement (3) is incorrect because hydrogen gas, not oxygen gas, is evolved at the cathode.

Award 1 mark for the correct option A.

According to the Arrhenius equation, \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). Therefore, the slope \(m = -\frac{E_a}{R}\). Thus, \(E_a = -\text{slope} \times R = -(-1.20 \times 10^4\text{ K}) \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 9.97 \times 10^4\text{ J mol}^{-1} = 99.7\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option B.

Propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) molecules can form intermolecular hydrogen bonds, which are much stronger than the dipole-dipole forces in methoxyethane and propanal, and the Van der Waals' forces in butane. Since all these compounds have similar relative molecular masses, propan-1-ol has the highest boiling point.

Award 1 mark for the correct option A.

Metal X is highly reactive as it reacts with cold water. Metal Y is moderately reactive as its oxide can be reduced by carbon but does not react with steam. Metal Z is very unreactive as it does not react with acid and its oxide is reduced by heating alone. Thus, the decreasing order of reactivity is X > Y > Z.

Award 1 mark for the correct option A.

In reaction C, \(\text{Fe}^{2+}\) is oxidized to \(\text{Fe}^{3+}\) (oxidation number increases from +2 to +3). Therefore, \(\text{Fe}^{2+}\) acts as a reducing agent. In other options, the underlined substances are reduced, so they act as oxidizing agents.

Award 1 mark for the correct option C.

As the forward reaction is exothermic (\(\Delta H < 0\)), increasing the temperature shifts the equilibrium position to the left (endothermic direction) to absorb heat. Therefore, the yield of \(\text{SO}_3(g)\) decreases and the value of \(K_c\) decreases. The rate of the forward reaction increases because of the higher temperature. The total pressure of the mixture increases due to both the higher temperature and the increase in the total number of gas molecules (shift to the side with 3 moles of gas instead of 2).

Award 1 mark for the correct option B.

Since X has the molecular formula \(C_4H_8O\), its degree of unsaturation is 1. The facts that X does not react with \(NaHCO_3\)(aq) rules out carboxylic acids. It does not react with acidified \(K_2Cr_2O_7\)(aq), which rules out primary/secondary alcohols (such as but-3-en-1-ol and cyclobutanol) and aldehydes. It rapidly decolourises \(Br_2\) in \(CH_2Cl_2\) in the dark, which indicates the presence of a C=C double bond, ruling out butanone and cyclobutanol. Ethoxyethene (\(CH_2=CH-O-CH_2CH_3\)) is an ether containing a C=C double bond. Ethers do not react with acidified \(K_2Cr_2O_7\), and the C=C bond reacts with bromine via addition. Thus, X is ethoxyethene.

Correct option: C (1 mark).

According to Hess's Law, the formation of propene is given by:\n\(3C(s, \text{graphite}) + 3H_2(g) \rightarrow C_3H_6(g)\)\n\(\Delta H^{\theta}_f [C_3H_6(g)] = 3 \Delta H^{\theta}_c [C(s, \text{graphite})] + 3 \Delta H^{\theta}_c [H_2(g)] - \Delta H^{\theta}_c [C_3H_6(g)]\)\n\(\Delta H^{\theta}_f [C_3H_6(g)] = 3(-393.5) + 3(-285.8) - (-2058.0) = -1180.5 - 857.4 + 2058.0 = +20.1 \text{ kJ mol}^{-1}\).

Correct option: B (1 mark).

Statement (1) is correct: PLA contains ester linkages, which can undergo hydrolysis catalyzed by acids, bases, or enzymes in the environment, making it biodegradable. Statement (2) is correct: PLA consists of linear chains without covalent cross-links, so it softens upon heating and can be reshaped (thermoplastic). Statement (3) is correct: The monomer is lactic acid, \(CH_3CH(OH)COOH\). The second carbon is bonded to four different groups (\(-H\), \(-CH_3\), \(-OH\), and \(-COOH\)), hence it is chiral.

Correct option: D (1 mark).

Assume 100 g of the oxide. Mass of Cl = 42.5 g, Mass of O = 100 - 42.5 = 57.5 g.\nNumber of moles of Cl = \(42.5 / 35.5 = 1.20 \text{ mol}\).\nNumber of moles of O = \(57.5 / 16.0 = 3.59 \text{ mol}\).\nRatio of moles of Cl to O = \(1.20 : 3.59 \approx 1 : 3\).\nThus, the empirical formula of the oxide is \(ClO_3\).

Correct option: C (1 mark).

In the electrolysis of concentrated \(NaCl\)(aq), at the anode, \(Cl^-\) ions are preferentially discharged over \(OH^-\) ions due to their high concentration, forming \(Cl_2(g)\). At the cathode, \(H^+(aq)\) ions (from water self-ionization) are preferentially discharged over \(Na^+(aq)\) ions because \(H^+\) is a stronger oxidizing agent, forming \(H_2(g)\). As \(H^+(aq)\) is discharged, excess \(OH^-(aq)\) accumulates around the cathode, causing the local pH to increase.

Correct option: B (1 mark).

From the Arrhenius equation: \(\ln k = -\frac{E_a}{R} \left(\frac{1}{T}\right) + \ln A\).\nThus, the slope of the plot of \(\ln k\) against \(\frac{1}{T}\) is equal to \(-\frac{E_a}{R}\).\nGiven slope = \(-1.20 \times 10^4 \text{ K}\):\n\(-\frac{E_a}{8.31} = -1.20 \times 10^4\)\n\(E_a = 1.20 \times 10^4 \times 8.31 = 9.972 \times 10^4 \text{ J mol}^{-1} \approx 99.7 \text{ kJ mol}^{-1}\).

Correct option: A (1 mark).

Statement (1) is correct: Transition metals have partly filled d-orbitals, allowing them to lose different numbers of d-electrons and exhibit variable oxidation states (e.g., \(Fe^{2+}\) and \(Fe^{3+}\)). Statement (2) is correct: Splitting of d-orbitals in transition metal complex ions allows d-d electronic transitions by absorbing visible light, rendering most aqueous transition metal ions coloured. Statement (3) is correct: Due to their ability to adopt multiple oxidation states and form complexes, transition metals and their compounds can act as homogeneous or heterogeneous catalysts by providing alternative reaction pathways with lower activation energies.

Correct option: D (1 mark).

All three compounds have similar molecular masses (propan-1-ol = 60, propanal = 58, butane = 58), so the strength of their intermolecular forces determines the boiling point. (1) Propan-1-ol molecules are held together by hydrogen bonds, which are the strongest. (2) Propanal molecules are polar and held together by permanent dipole-permanent dipole attractions (van der Waals' forces). (3) Butane molecules are non-polar and held together only by weak instantaneous dipole-induced dipole forces. Thus, the order of decreasing boiling point is Propan-1-ol > Propanal > Butane, i.e., (1) > (2) > (3).

Correct option: A (1 mark).

When the volume is halved, the concentration of both \(NO_2(g)\) and \(N_2O_4(g)\) immediately doubles, causing the brown colour to become darker instantly. According to Le Chatelier's Principle, the decrease in volume (increase in pressure) shifts the equilibrium to the right (the side with fewer gaseous moles, i.e., forming more \(N_2O_4\) and consuming \(NO_2\)). Thus, the concentration of \(NO_2(g)\) gradually decreases, making the brown colour fade slightly. However, according to Le Chatelier's Principle, the net effect of the initial concentration increase cannot be completely neutralized, so the final equilibrium concentration of \(NO_2(g)\) is still higher than its initial value. Therefore, the brown colour remains darker than the original colour.

Correct option: C (1 mark).

Statement (1) is correct: Zinc is more reactive than copper, so it is oxidized to \(Zn^{2+}(aq)\) at the anode. Since \(Zn^{2+}(aq)\) is a weaker oxidizing agent than \(Cu^{2+}(aq)\), it is not reduced at the cathode and remains in the solution. Statement (2) is correct: Silver and gold are less reactive than copper, so they are not oxidized at the anode potential and fall to the bottom as anode slime. Statement (3) is correct: At the cathode, only \(Cu^{2+}\) is reduced to copper. At the anode, both zinc and copper are oxidized. Since part of the current is used to oxidize zinc, fewer \(Cu^{2+}\) ions are produced at the anode than are consumed at the cathode. Hence, the concentration of \(Cu^{2+}(aq)\) in the electrolyte decreases gradually.

Correct option: D (1 mark).

Since the compound \(W\) (\(\text{C}_4\text{H}_8\text{O}_2\)) is hydrolyzed by dilute \(\text{NaOH}\) under reflux to give two products, it must be an ester. One of the products, \(P\), can be oxidized to a ketone, which means \(P\) must be a secondary alcohol. Among the options, only 1-methylethyl methanoate (isopropyl methanoate) hydrolyzes to give propan-2-ol (a secondary alcohol) and methanoate ions. Propan-2-ol is oxidized to propanone (a ketone). Propyl methanoate gives propan-1-ol (a primary alcohol, which oxidizes to an aldehyde/carboxylic acid).

Award 1 mark for the correct option (B).

The target equation for the formation of ethyne is: \(2\text{C(graphite)} + \text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_2\text{(g)}\). Using Hess's Law: \(\Delta H_{\text{f}}^\theta[\text{C}_2\text{H}_2\text{(g)}] = 2 \Delta H_1 + \Delta H_2 - \frac{1}{2}\Delta H_3\). \(\Delta H_{\text{f}}^\theta = 2(-393.5) + (-285.8) - \frac{1}{2}(-2598.8) = -787.0 - 285.8 + 1299.4 = +226.6\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option (A).

Statement (1) is correct: The repeating unit of the polymer is \(-\text{CH}_2-\text{CH(Cl)}-\), which comes from the monomer chloroethene (\(\text{CH}_2=\text{CHCl}\)). Statement (2) is correct: This is polyvinyl chloride (PVC), which is a thermoplastic and can be softened by heating. Statement (3) is incorrect: Chloroethene has two hydrogen atoms attached to one of the double-bonded carbon atoms, so it does not exhibit cis-trans isomerism.

Award 1 mark for the correct option (A).

To determine the molecular shape, we use VSEPR theory: \(\text{BF}_3\) is trigonal planar (3 bonding pairs, 0 lone pairs). \(\text{H}_3\text{O}^+\) has a central O atom with 5 valence electrons (6 minus 1 for positive charge), leaving 3 single bonds with H and 1 lone pair, which gives a trigonal pyramidal shape. \(\text{CO}_3^{2-}\) is trigonal planar. \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs, giving a tetrahedral shape.

Award 1 mark for the correct option (B).

During the electrolysis of concentrated \(\text{NaCl(aq)}\): At the cathode (-), \(\text{H}^+\text{(aq)}\) ions (from water dissociation) are preferentially discharged over \(\text{Na}^+\text{(aq)}\) ions to form hydrogen gas: \(2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}\). This removal of \(\text{H}^+\) leaves an excess of \(\text{OH}^-\)_ ions around the cathode, causing the local pH to increase. At the anode (+), \(\text{Cl}^-\text{(aq)}\) ions are preferentially discharged over \(\text{OH}^-\text{(aq)}\) due to their high concentration, producing chlorine gas: \(2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-\).

Award 1 mark for the correct option (C).

According to the Arrhenius equation, \(\ln k = -\frac{E_{\text{a}}}{R} \cdot \frac{1}{T} + \ln A\). The slope of the plot of \(\ln k\) against \(\frac{1}{T}\) is equal to \(-\frac{E_{\text{a}}}{R}\). Thus, \(-\frac{E_{\text{a}}}{R} = -1.20 \times 10^4\text{ K}\) which gives \(E_{\text{a}} = 1.20 \times 10^4\text{ K} \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 99720\text{ J mol}^{-1} = 99.72\text{ kJ mol}^{-1}\). Therefore, the activation energy of the reaction is approximately \(99.7\text{ kJ mol}^{-1}\).

Award 1 mark for the correct option (A).

(a) Heat absorbed by water, \( Q = m c \Delta T = 150.0\text{ g} \times 4.2\text{ J g}^{-1}\text{ K}^{-1} \times 32.5\text{ K} = 20475\text{ J} = 20.475\text{ kJ} \).\nNumber of moles of propan-1-ol burned = \( \frac{0.90}{60.0} = 0.015\text{ mol} \).\nEnthalpy change of combustion, \( \Delta H = -\frac{20.475}{0.015} = -1365\text{ kJ mol}^{-1} \).\n(b) Heat lost to the surroundings; Incomplete combustion of propan-1-ol; Heat absorbed by the copper calorimeter was not accounted for.\n(c) Use a wind shield to minimize heat loss to surroundings / use a bomb calorimeter / insulate the calorimeter.

Part (a) [3 marks]:\n- Correct calculation of heat absorbed (20.475 kJ) [1M]\n- Correct calculation of moles of propan-1-ol (0.015 mol) [1M]\n- Correct calculation of enthalpy change with negative sign and units (-1365 kJ mol^-1) [1M]\nPart (b) [2 marks]:\n- State any two valid reasons (e.g., heat lost to surrounding, incomplete combustion, heat capacity of calorimeter ignored) [1M each]\nPart (c) [1 mark]:\n- Suggest a valid improvement (e.g., use draft shield, insulate, use bomb calorimeter) [1M]

(a) Step 1 (cyclohexanone to cyclohexanol): \( \text{NaBH}_4 \) in methanol at room temperature. Step 2 (cyclohexanol to cyclohexene): Excess concentrated \( \text{H}_2\text{SO}_4 \) heating at around \( 170 ^\circ\text{C} \) (or reflux).\n(b) Elimination (or dehydration).\n(c) Reagent: bromine in tetrachloromethane (or \( \text{Br}_2\text{ in CH}_2\text{Cl}_2 \)) at room temperature in the dark. Color change: orange-red/brown to colorless.

Part (a) [3 marks]:\n- Correct reagent for reduction (e.g., NaBH4 in alcohol/water) [1M]\n- Correct reagent for dehydration (conc. H2SO4 or conc. H3PO4) [1M]\n- Heating/reflux condition for dehydration [1M]\nPart (b) [1 mark]:\n- Elimination / Dehydration [1M]\nPart (c) [3 marks]:\n- Correct reagent (Br2 in organic solvent like CH2Cl2) [1M]\n- Condition (room temperature, dark/avoid direct sunlight) [1M]\n- Color change from orange/brown to colorless [1M]

(a) Repeating unit: \( \text{[-O-CH(CH}_3\text{)-CO-]} \)\n(b) PLA contains ester linkages (\( \text{-COO-} \)) along its main polymer chain, which are polar and susceptible to hydrolysis by enzymes or water in the environment. Polyethene has a non-polar carbon-carbon (\( \text{C-C} \)) backbone which is chemically inert and cannot be easily hydrolyzed or broken down by microorganisms.\n(c) PLA is derived from renewable resources (such as corn starch) rather than fossil fuels / It has a lower carbon footprint during raw material production.

Part (a) [2 marks]:\n- Correct connectivity of the ester repeating unit with open bonds shown at both ends [2M] (deduct 1M for missing methyl side group or incorrect backbone)\nPart (b) [3 marks]:\n- PLA has polar ester links, susceptible to hydrolysis/attack by microorganisms/enzymes [1M]\n- Polyethene has a non-polar strong C-C single bond backbone [1M]\n- Polyethene is chemically inert and resistant to microbial attack [1M]\nPart (c) [1 mark]:\n- Made from renewable resources / reduces carbon footprint [1M]

(a) For \( \text{NF}_3 \), N has 1 lone pair and 3 bond pairs with F (each F has 3 lone pairs). For \( \text{BF}_3 \), B has 3 bond pairs with F (each F has 3 lone pairs), and B has an incomplete octet (6 valence electrons).\n(b) \( \text{NF}_3 \) shape: Trigonal pyramidal, bond angle: \( \approx 102.5^\circ \) (accept \( 101^\circ - 107^\circ \)). \( \text{BF}_3 \) shape: Trigonal planar, bond angle: \( 120^\circ \).\nExplanation: In \( \text{NF}_3 \), the central N atom has 4 electron domains (3 bonding pairs and 1 lone pair). Due to lone pair-bond pair repulsion being greater than bond pair-bond pair repulsion, the bond angle is compressed to form a trigonal pyramidal shape. In \( \text{BF}_3 \), the central B atom has only 3 electron domains (all 3 are bonding pairs), which repel each other minimally at \( 120^\circ \) in a flat plane.\n(c) \( \text{NF}_3 \) is polar because the dipoles of the polar \( \text{N-F} \) bonds do not cancel out due to the asymmetrical trigonal pyramidal shape. \( \text{BF}_3 \) is non-polar because the highly symmetrical trigonal planar shape causes the dipoles of the three polar \( \text{B-F} \) bonds to cancel out completely.

Part (a) [1 mark]:\n- Correct electron diagrams showing valence shell electrons for both molecules (outermost shell only is acceptable) [1M]\nPart (b) [3 marks]:\n- Correct shapes and angles for both: NF3 (trigonal pyramidal, < 109.5) and BF3 (trigonal planar, 120) [1M]\n- State that N has 3 bond pairs and 1 lone pair, while B has 3 bond pairs and 0 lone pairs [1M]\n- State lone pair-bond pair repulsion > bond pair-bond pair repulsion resulting in the shape differences [1M]\nPart (c) [2 marks]:\n- Explain NF3 is polar due to asymmetrical shape / polar bonds not cancelling out [1M]\n- Explain BF3 is non-polar due to symmetrical shape / polar bonds cancelling out [1M]

(a) Diagram should show: a DC power source, two carbon/graphite electrodes immersed in a beaker of concentrated NaCl(aq), inverted test tubes filled with electrolyte over the electrodes to collect gases. Labelling of anode (+), cathode (-), electrolyte, and gases collected.\n(b) Anode reaction: \( 2\text{Cl}^-(\text{aq}) \rightarrow \text{Cl}_2(\text{g}) + 2\text{e}^- \)\nCathode reaction: \( 2\text{H}_2\text{O}(\text{l}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq}) \) (or \( 2\text{H}^+(\text{aq}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) \))\n(c) At the cathode, \( \text{H}^+(\text{aq}) \) ions (from self-ionization of water) are preferentially discharged, leaving behind \( \text{OH}^-(\text{aq}) \) ions in the solution. This causes the concentration of \( \text{OH}^- \) ions near the cathode to increase, making the solution alkaline, so the pH increases.\n(d) Chlorine gas: Manufacture of PVC, bleaching agents, sterilizing swimming pool water/drinking water, or production of hydrochloric acid.

Part (a) [2 marks]:\n- Well-drawn and complete diagram of electrolysis setup [1M]\n- Complete and correct labels (electrodes, electrolyte, anode/cathode polarity) [1M]\nPart (b) [2 marks]:\n- Correct anode half equation [1M]\n- Correct cathode half equation [1M]\nPart (c) [2 marks]:\n- State pH increases/becomes alkaline [1M]\n- Explain due to discharge of H+ ions (from H2O), leaving excess OH- ions in the solution [1M]\nPart (d) [1 mark]:\n- State any correct industrial use of chlorine [1M]

(a) Let \( k_1 = 1.3 \times 10^{-3} \), \( T_1 = 600\text{ K} \).\nLet \( k_2 = 2.3 \times 10^{-2} \), \( T_2 = 650\text{ K} \).\n\( \ln\left(\frac{2.3 \times 10^{-2}}{1.3 \times 10^{-3}}\right) = -\frac{E_a}{8.31}\left(\frac{1}{650} - \frac{1}{600}\right) \)\n\( \ln(17.692) = -\frac{E_a}{8.31}\left(0.0015385 - 0.0016667\right) \)\n\( 2.873 = -\frac{E_a}{8.31}\left(-1.282 \times 10^{-4}\right) \)\n\( 2.873 = E_a \times 1.543 \times 10^{-5} \)\n\( E_a = \frac{2.873}{1.543 \times 10^{-5}} = 186195\text{ J mol}^{-1} = 186\text{ kJ mol}^{-1} \).\n(b) A catalyst provides an alternative reaction pathway with a lower activation energy. Thus, a larger fraction of reactant molecules have kinetic energy greater than or equal to the lower activation energy, leading to a higher frequency of effective collisions.

Part (a) [4 marks]:\n- Correct substitution of values into Arrhenius equation [1M]\n- Calculation of the natural log ratio (approx. 2.87) [1M]\n- Calculation of temperature inverse difference (approx. -1.28 x 10^-4) [1M]\n- Correct final answer in kJ mol^-1 (accept 185 to 186) [1M]\nPart (b) [2 marks]:\n- State catalyst provides alternative pathway with lower activation energy [1M]\n- More molecules have energy >= Ea, increasing frequency of effective collisions [1M]

(a) \( \text{N}_2(\text{g}) + 2\text{H}_2(\text{g}) \rightarrow \text{N}_2\text{H}_4(\text{l}) \)\n(b) Cycle:\nReactants: \( \text{N}_2(\text{g}) + 2\text{H}_2(\text{g}) + 2\text{O}_2(\text{g}) \)\nPath 1 (direct combustion of H2): \( \text{N}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g}) \) with \( \Delta H = 2 \times (-286) = -572\text{ kJ} \).\nPath 2 (formation of hydrazine, followed by its combustion):\n\( \text{N}_2(\text{g}) + 2\text{H}_2(\text{g}) \rightarrow \text{N}_2\text{H}_4(\text{l}) \) (with enthalpy change \( \Delta H_f^
ormal_standard[\text{N}_2\text{H}_4(\text{l})] \))\nfollowed by: \( \text{N}_2\text{H}_4(\text{l}) + \text{O}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \) (with enthalpy change \( -622\text{ kJ} \)).\nTherefore, by Hess's Law: \( 2 \times (-286) = \Delta H_f^
ormal_standard[\text{N}_2\text{H}_4(\text{l})] + (-622) \)\n(c) \( -572 = \Delta H_f^
ormal_standard[\text{N}_2\text{H}_4(\text{l})] - 622 \)\n\( \Delta H_f^
ormal_standard[\text{N}_2\text{H}_4(\text{l})] = -572 + 622 = +50\text{ kJ mol}^{-1} \).

Part (a) [1 mark]:\n- Correct balanced equation with state symbols [1M]\nPart (b) [3 marks]:\n- Correctly constructed energy cycle showing reactants, intermediates, and products with arrows [2M] (deduct 1M for incorrect stoichiometry or state symbols)\n- Correct labels of enthalpy terms on arrows [1M]\nPart (c) [2 marks]:\n- Setup equation using Hess's Law [1M]\n- Correct calculation of enthalpy of formation (+50 kJ mol^-1) including '+' sign and unit [1M]

(a) Propanone: \( \text{CH}_3\text{COCH}_3 \);\nPropanal: \( \text{CH}_3\text{CH}_2\text{CHO} \);\nProp-2-en-1-ol: \( \text{CH}_2\text{=CHCH}_2\text{OH} \).\n(b) Reagent: Tollens' reagent (or Fehling's solution / acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \)).\nObservation: Propanal forms a silver mirror with Tollens' reagent, whereas propanone shows no observable change.\n(c) Reagent: Bromine in organic solvent (e.g. \( \text{Br}_2\text{ in CH}_2\text{Cl}_2 \)) in the dark.\nObservation: Prop-2-en-1-ol decolorizes the orange-red bromine solution immediately, whereas propanone and propanal do not show immediate decolorization in the dark.

Part (a) [3 marks]:\n- Correct structure of propanone [1M]\n- Correct structure of propanal [1M]\n- Correct structure of prop-2-en-1-ol [1M]\nPart (b) [2 marks]:\n- Correct reagent (e.g., Tollens' or Fehling's) [1M]\n- Correct observations for both (silver mirror / red ppt for propanal, no change for propanone) [1M]\nPart (c) [2 marks]:\n- Correct reagent (Br2 in organic solvent in dark) [1M]\n- Correct observation (decolorizes for prop-2-en-1-ol, no immediate decolorization for others) [1M]

(a) Silicon dioxide has a much higher melting point than carbon dioxide.\nSilicon dioxide has a giant covalent structure. To melt it, strong covalent bonds between silicon and oxygen atoms must be broken, which requires a huge amount of energy.\nCarbon dioxide has a simple molecular structure. To melt/sublime it, only weak intermolecular forces (van der Waals' forces) between carbon dioxide molecules need to be overcome, which requires very little energy.\n(b) Silicon dioxide cannot conduct electricity in the molten state. All valence electrons are localized in strong covalent bonds, and there are no delocalized electrons or free-moving ions to carry the electric charge.

Part (a) [4 marks]:\n- Correct comparison of melting points (SiO2 high, CO2 low) [1M]\n- Identify giant covalent structure for SiO2 and explain melting requires breaking strong covalent bonds [1M]\n- Identify simple molecular structure for CO2 [1M]\n- Explain melting/sublimation of CO2 only requires overcoming weak van der Waals' forces [1M]\nPart (b) [2 marks]:\n- Correctly states SiO2 does not conduct electricity in molten state [1M]\n- Explain due to absence of delocalized electrons or free-moving ions [1M]

(a) Monomers are benzene-1,4-dicarboxylic acid (\( \text{HOOC-C}_6\text{H}_4\text{-COOH} \)) and benzene-1,4-diamine (\( \text{H}_2\text{N-C}_6\text{H}_4\text{-NH}_2 \)).\n(b) Condensation polymerisation.\n(c) Kevlar chains are highly aligned and planar. The carbonyl oxygen (with a partial negative charge) from one chain can form strong hydrogen bonds with the hydrogen atom (with a partial positive charge) attached to the nitrogen in an adjacent chain. The extensive network of strong hydrogen bonds between the closely packed chains binds them tightly together, resisting separation under high tension.

Part (a) [3 marks]:\n- Correct structure of benzene-1,4-dicarboxylic acid (or its diacyl chloride) [1.5M]\n- Correct structure of benzene-1,4-diamine [1.5M]\nPart (b) [1 mark]:\n- Condensation polymerisation [1M]\nPart (c) [3 marks]:\n- Identify hydrogen bonding as the key intermolecular force [1M]\n- Explain hydrogen bonds form between C=O of one chain and N-H of adjacent chain [1M]\n- Mention that the linear/planar structures allow close packing and an extensive network of hydrogen bonds [1M]

(a) Use a measuring cylinder to transfer \(50.0\text{ cm}^3\) of hydrochloric acid into a polystyrene cup. Measure and record the initial temperature of the acid. Weigh \(3.40\text{ g}\) of \(LiHCO_3(s)\) accurately. Add the solid to the acid, stir continuously, and record the lowest temperature reached. To minimize heat loss, use a lid on the polystyrene cup or place the cup inside a beaker for extra insulation.

(b) Moles of \(LiHCO_3\) = \(\frac{3.40}{68.0} = 0.050\text{ mol}\)
Heat absorbed by the solution, \(q = m c \Delta T = 50.0\text{ g} \times 4.2\text{ J g}^{-1}\ \text{K}^{-1} \times (21.5 - 17.2)\text{ K} = 903\text{ J} = 0.903\text{ kJ}\)
Since temperature decreases, the reaction is endothermic.
\(\Delta H_2 = +\frac{0.903\text{ kJ}}{0.050\text{ mol}} = +18.06\text{ kJ mol}^{-1}\) (or \(+18.1\text{ kJ mol}^{-1}\)).

(c) According to Hess's Law, the decomposition reaction is represented by:
\(2LiHCO_3(s) \rightarrow Li_2CO_3(s) + CO_2(g) + H_2O(l)\)
This can be constructed from: \(2 \times \text{Reaction (2)} - \text{Reaction (1)}\)
\(\Delta H_{dec} = 2\Delta H_2 - \Delta H_1 = 2(+18.06) - (-25.0) = +36.12 + 25.0 = +61.12\text{ kJ mol}^{-1}\) (accept \(+61.1\text{ kJ mol}^{-1}\) or \(+61.2\text{ kJ mol}^{-1}\)).

(a) Description: Measure initial temperature of acid, add solid, stir and record lowest temperature (2 marks; 1 mark for initial/lowest temp, 1 mark for stirring/mixing). Precaution: Use a lid / wrap with cotton wool / put inside a beaker (1 mark).
(b) Moles of \(LiHCO_3\) = 0.050 mol (0.46 mark).
\(q = 50.0 \times 4.2 \times 4.3 = 903\text{ J}\) (1 mark).
\(\Delta H_2 = +18.1\text{ kJ mol}^{-1}\) (1 mark for correct value and positive sign).
(c) \(\Delta H_{dec} = 2(+18.06) - (-25.0) = +61.1\text{ kJ mol}^{-1}\) (1 mark; accept calculations based on student's value in part b).

(a) Condensation polymerisation.
Repeating unit structure:
\(-[O-CH_2-CO-O-CH(CH_3)-CO]-\) (or its reversed or structural equivalent, showing clear ester linkages with open bonds at both ends).

(b) PLGA contains polar ester linkages (\(-CO-O-\)) in its polymer backbone, which can be readily hydrolysed by water or broken down by microbial enzymes. Polyethene consists of a long hydrocarbon chain with only non-polar C-C and C-H bonds, which are highly stable and resistant to enzymatic attack or hydrolysis.

(c) Lactic acid contains a hydrophobic methyl group (\(-CH_3\)) as a side chain, whereas glycolic acid has only hydrogen atoms. A higher proportion of lactic acid increases the overall hydrophobicity of the copolymer, which repels water molecules and decreases the rate of water penetration into the polymer, thereby slowing down the rate of ester hydrolysis (degradation).

(a) Condensation polymerisation (0.46 mark). Correct structural formula of the repeating unit with open bonds at both ends and correct ester linkages (2 marks; deduct 1 mark for missing open bonds or minor errors).
(b) PLGA has ester linkages which can be hydrolysed/enzymatically degraded (1 mark). Polyethene has non-polar C-C and C-H bonds which are inert/resistant to hydrolysis (1 mark).
(c) Lactic acid has a hydrophobic methyl group (1 mark). Higher proportion of lactic acid increases hydrophobicity / decreases water absorption, reducing the rate of hydrolysis (1 mark).

(a) Activation energy is the minimum energy that colliding reactant particles must possess in order for a chemical reaction to occur.

(b) Using the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\)
\(\ln\left(\frac{1.50 \times 10^{-3}}{3.46 \times 10^{-5}}\right) = -\frac{E_a}{8.31} \left(\frac{1}{328} - \frac{1}{298}\right)\)
\(\ln(43.35) = -\frac{E_a}{8.31} \left(0.003049 - 0.003356\right)\)
\(3.769 = -\frac{E_a}{8.31} \left(-3.07 \times 10^{-4}\right)\)
\(3.769 = E_a \times 3.69 \times 10^{-5}\)
\(E_a = 1.02 \times 10^5\text{ J mol}^{-1} = 102\text{ kJ mol}^{-1}\) (accept \(101.5\) to \(102.5\text{ kJ mol}^{-1}\)).

(c) A sketched diagram showing:
- Y-axis: Potential Energy (or Energy); X-axis: Progress of reaction (or Reaction coordinate).
- Reactants at higher energy level than products (exothermic profile).
- A curve rising to a peak (transition state) and falling to products.
- Double-headed arrows clearly indicating:
1. \(E_a\) from reactant level to peak.
2. \(\Delta H\) from reactant level to product level (pointing downwards or labelled with negative value).

(a) Definition: Minimum energy required for a successful collision/reaction to occur (1 mark).
(b) Substitution of values into Arrhenius equation (1 mark).
Calculation of \(\ln(k_2/k_1)\) and \(\Delta(1/T)\) correctly (1 mark).
Final answer \(102\text{ kJ mol}^{-1}\) (or \(102.1\text{ kJ mol}^{-1}\)) with unit (1.46 marks; deduct 0.46 marks for incorrect unit or missing unit).
(c) Exothermic profile with correct labels for axes (1 mark). Correct positioning and labeling of \(E_a\) and \(\Delta H\) with arrows (1 mark).

**Part (a)**
Use the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\) or \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
Substitute the values:
\(\ln\left(\frac{1.12 \times 10^{-2}}{1.45 \times 10^{-3}}\right) = \frac{E_a}{8.31} \left(\frac{1}{523} - \frac{1}{573}\right)\)
\(\ln(7.7241) = \frac{E_a}{8.31} \left(1.9120 \times 10^{-3} - 1.7452 \times 10^{-3}\right)\)
\(2.0444 = \frac{E_a}{8.31} \left(1.6684 \times 10^{-4}\right)\)
\(E_a = \frac{2.0444 \times 8.31}{1.6684 \times 10^{-4}} = 101827\text{ J mol}^{-1} = 101.8\text{ kJ mol}^{-1}\) (or \(102\text{ kJ mol}^{-1}\)).

**Part (b)**
(i) Increasing the pressure has no effect on the equilibrium yield of dimethyl ether. According to Le Chatelier's Principle, pressure shifts the equilibrium only if there is a difference in the number of moles of gaseous reactants and gaseous products. Here, there are 2 moles of gaseous reactants (\(2\text{CH}_3\text{OH}\)) and 2 moles of gaseous products (\(1\text{CH}_3\text{OCH}_3 + 1\text{H}_2\text{O}\)).
(ii) From a kinetic viewpoint, a higher temperature increases the reaction rate, ensuring that the chemical equilibrium is reached within an acceptable time frame in an industrial reactor.
From a thermodynamic viewpoint, since the reaction is exothermic, a higher temperature shifts the equilibrium to the reactant side, decreasing the equilibrium yield of dimethyl ether.
Therefore, a compromise temperature of \(250 - 300^\circ\text{C}\) is used to balance reaction rate and equilibrium yield.

**Part (c)**
(i) The catalyst provides an alternative reaction pathway with a lower activation energy, thereby increasing the rate of both forward and reverse reactions.
(ii) Equation: \(2\text{CH}_3\text{OH} \rightarrow \text{CH}_3\text{OCH}_3 + \text{H}_2\text{O}\)
Molar mass of reactant (\(2\text{CH}_3\text{OH}\)) = \(2 \times [12.0 + (1.0 \times 4) + 16.0] = 2 \times 32.0 = 64.0\text{ g mol}^{-1}\)
Molar mass of desired product (\(\text{CH}_3\text{OCH}_3\)) = \((12.0 \times 2) + (1.0 \times 6) + 16.0 = 46.0\text{ g mol}^{-1}\)
Atom Economy = \(\frac{\text{Molar mass of desired product}}{\text{Total molar mass of reactants}} \times 100\%\)
Atom Economy = \(\frac{46.0}{64.0} \times 100\% = 71.9\%\).
(iii) Any two of the following:
- **Catalysis**: Using a catalyst (alumina) rather than stoichiometric reagents reduces energy requirements and increases selectivity.
- **Use of Renewable Feedstocks**: Methanol can be produced from biomass (biomethanol) rather than fossil resources.
- **Design for Energy Efficiency**: The process is run at moderate temperatures, and waste heat from this exothermic reaction can be recovered.

**Part (a) (6 marks)**
- Correct Arrhenius equation formulation (1)
- Correct substitution of $k_1, k_2, T_1, T_2$ (1)
- Correct evaluation of $\ln(k_2/k_1)$ as 2.04 and $(1/T_1 - 1/T_2)$ as $1.67 \times 10^{-4}$ (1)
- Calculation of $E_a$ in $\text{J mol}^{-1}$ as 101827 (1)
- Correct conversion to $\text{kJ mol}^{-1}$ (101.8 or 102) (1)
- Correct unit (\(\text{kJ mol}^{-1}\)) and positive sign (1)

**Part (b) (6 marks)**
(i)
- State "No effect" on the yield (1)
- Explain that the number of moles of gaseous reactants equals gaseous products (1)
- Cite Le Chatelier's Principle or constant concentration ratio (1)
(ii)
- Explain kinetic reason: Higher temperature increases rate / effective collisions (1)
- Explain thermodynamic reason: Exothermic reaction, so higher temperature decreases equilibrium yield (1)
- State that a compromise temperature balances yield and rate (1)

**Part (c) (8 marks)**
(i)
- Alternative pathway (1)
- Lower activation energy (1)
(ii)
- Correct calculation of molar masses of reactant ($64.0$) and product ($46.0$) (1)
- Division and multiplying by 100% (1)
- Correct final answer: $71.9\%$ (accept $71.88\%$ - $72.0\%$) (1)
(iii)
- Identify any 2 appropriate green chemistry principles (1+1)
- Briefly explain how each applies to the scenario (1)

**Part (a)**
(i) The sharp peak at \(1735\text{ cm}^{-1}\) indicates a carbonyl group (\(\text{C=O}\) stretch). The absence of broad absorption bands at \(2500-3300\text{ cm}^{-1}\) (\(\text{O-H}\) stretch of carboxylic acid) and \(3230-3670\text{ cm}^{-1}\) (\(\text{O-H}\) stretch of alcohol) rules out carboxylic acids and alcohols. Therefore, \(W\) must contain an ester (\(\text{-COO-}\)) functional group.
(ii) The molecular ion peak at \(m/z = 88\) is consistent with the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) (Molar mass = \(88.0\text{ g mol}^{-1}\)).
The fragment peak at \(m/z = 57\) corresponds to the loss of a neutral radical of mass \(88 - 57 = 31\) units.
A mass loss of 31 is characteristic of a methoxy group (\(\cdot\text{OCH}_3\)). This shows that \(W\) is a methyl ester (\(\text{R-COOCH}_3\)).
Since the total molecular formula is \(\text{C}_4\text{H}_8\text{O}_2\), the remaining group \(\text{R}\) must be an ethyl group (\(-\text{CH}_2\text{CH}_3\), mass = 29).
Therefore, \(W\) is methyl propanoate. Its structural formula is \(\text{CH}_3\text{CH}_2\text{COOCH}_3\).

**Part (b)**
(i) Colorimetry is based on Beer-Lambert's Law, which states that the absorbance (\(A\)) of a coloured solution is directly proportional to its concentration (\(c\)) for a given path length.
(ii) Aqueous iron(II) ions are pale green and have a very low molar absorptivity (the colour is too weak to measure accurately). Reacting \(\text{Fe}^{2+}\) with 1,10-phenanthroline forms an intensely coloured red-orange complex, which greatly increases the sensitivity and accuracy of the absorbance measurement.
(iii) First, prepare a series of solutions of known \(\text{Fe}^{2+}\) concentrations (standard solutions). Measure the absorbance of each standard solution at the wavelength of maximum absorption (\(\lambda_{\text{max}}\)) using a colorimeter. Plot a graph of absorbance against concentration to obtain a straight calibration line passing through the origin.
(iv)
(1) Concentration of iron in Solution B = \(3.15\text{ mg dm}^{-3}\)
Volume of Solution B = \(100.0\text{ cm}^3 = 0.100\text{ dm}^3\)
Mass of iron in Solution B = \(3.15\text{ mg dm}^{-3} \times 0.100\text{ dm}^3 = 0.315\text{ mg}\).
This mass of iron is present in the \(10.0\text{ cm}^3\) aliquot of Solution A.
Since the total volume of Solution A is \(250.0\text{ cm}^3\), the total mass of iron in Solution A (and thus in the tablet) is:
\(0.315\text{ mg} \times \frac{250.0\text{ cm}^3}{10.0\text{ cm}^3} = 7.875\text{ mg}\) (or \(7.88\text{ mg}\)).

(2) Mass of the tablet = \(0.520\text{ g} = 520\text{ mg}\)
Percentage by mass of iron = \(\frac{7.875\text{ mg}}{520\text{ mg}} \times 100\% = 1.51\%\) (or \(1.514\%\)).

**Part (a) (7 marks)**
(i)
- Identify C=O at $1735\text{ cm}^{-1}$ (1)
- Explain that the absence of O-H peaks rules out acids/alcohols, concluding $W$ is an ester (1)
(ii)
- Relate $m/z = 88$ to formula $\text{C}_4\text{H}_8\text{O}_2$ (1)
- Deduce loss of 31 units represents loss of $\cdot\text{OCH}_3$ (1)
- Identify $W$ as a methyl ester, hence $\text{R} = \text{-C}_2\text{H}_5$ (1)
- Conclude name/structure: methyl propanoate (1)
- Draw correct structural formula $\text{CH}_3\text{CH}_2\text{COOCH}_3$ (1)

**Part (b) (13 marks)**
(i)
- State that absorbance is directly proportional to concentration (1)
- Refer to Beer-Lambert's Law / path length constant (1)
(ii)
- Mention that aqueous $\text{Fe}^{2+}$ is too pale for direct colorimetric analysis (1)
- Complexing agent forms intense colour to increase sensitivity/limit of detection (1)
(iii)
- Measure absorbance of standard solutions of known concentration (1)
- Measure at maximum absorption wavelength (1)
- Plot absorbance vs concentration to get a straight line passing through origin (1)
(iv)(1)
- Calculate mass of Fe in Solution B: $3.15 \times 0.100 = 0.315\text{ mg}$ (1)
- State dilution factor $250 / 10 = 25$ (1)
- Multiply to find mass in tablet: $0.315 \times 25 = 7.875\text{ mg}$ (1)
- Correct unit and significant figures (7.88 mg) (1)
(iv)(2)
- Correct expression: $7.875 / 520 \times 100\%$ (1)
- Correct calculation: $1.51\%$ (1)