2021 HKDSE Chemistry Answers & Marking Scheme

In option B, the reaction is NaCl(s) + H2SO4(l) -> NaHSO4(s) + HCl(g). The oxidation states of all elements remain unchanged: Na is +1, Cl is -1, H is +1, S is +6, and O is -2. Thus, it is a non-redox reaction. In A, chlorine undergoes disproportionation. In C, copper(II) nitrate decomposes to copper(II) oxide, nitrogen dioxide, and oxygen, where nitrogen is reduced and oxygen is oxidized. In D, carbon is oxidized to CO2 while nitric acid is reduced to NO2.

1 mark for the correct answer (B). 0 marks for incorrect options.

According to Hess's Law, the enthalpy change of formation of ethanol is given by: \(\Delta H_f^\theta[\text{C}_2\text{H}_5\text{OH(l)}] = 2 \times \Delta H_c^\theta[\text{C(s)}] + 3 \times \Delta H_c^\theta[\text{H}_2\text{(g)}] - \Delta H_c^\theta[\text{C}_2\text{H}_5\text{OH(l)}]\). Substituting the values: \(\Delta H_f^\theta = 2(-393.5) + 3(-285.8) - (-1367.3) = -787.0 - 857.4 + 1367.3 = -277.1\text{ kJ mol}^{-1}\).

1 mark for the correct answer (A). 0 marks for incorrect options.

Step 1: Heating propan-1-ol with excess concentrated sulfuric acid at \(170\ ^\circ\text{C}\) dehydrates it to form propene. Step 2: Acid-catalyzed hydration of propene (\(\text{H}_2\text{O}/\text{H}^+\)) yields propan-2-ol as the major product because the hydroxyl group adds to the secondary carbon according to Markovnikov's rule.

1 mark for the correct answer (A). 0 marks for incorrect options.

From clue 1, \(Y\) is more reactive than \(X\) because its oxide cannot be reduced by carbon, whereas \(X\)'s oxide can. From clue 2, \(X\) is more reactive than \(Z\) because \(X\) can displace \(Z\) from its nitrate solution. From clue 3, \(Y\) is highly reactive as it reacts with cold water. Therefore, the order of reactivity is \(Y > X > Z\).

1 mark for the correct answer (A). 0 marks for incorrect options.

Using the Arrhenius equation: \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\). Substituting the given values: \(\ln\left(\frac{8.0 \times 10^{-3}}{2.0 \times 10^{-3}}\right) = \frac{E_a}{8.31}\left(\frac{1}{300} - \frac{1}{320}\right)\) => \(\ln(4) = \frac{E_a}{8.31} \times \frac{20}{96000}\) => \(1.3863 = \frac{E_a}{8.31} \times \frac{1}{4800}\) => \(E_a = 1.3863 \times 8.31 \times 4800 = 55298\text{ J mol}^{-1} \approx 55.3\text{ kJ mol}^{-1}\).

1 mark for the correct answer (A). 0 marks for incorrect options.

Statement (1) is incorrect because the given reduction half-equation occurs at the cathode, not the anode. The anode reaction is the oxidation of hydrogen: \(\text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + 2e^-\). Statements (2) and (3) are correct: KOH(aq) is the standard electrolyte in alkaline fuel cells, and the only product of the overall cell reaction is water.

1 mark for the correct answer (D). 0 marks for incorrect options.

Total initial moles of \(\text{HCl} = 0.0500\text{ dm}^3 \times 0.500\text{ mol dm}^{-3} = 0.0250\text{ mol}\). Moles of \(\text{HCl}\) neutralized by \(\text{NaOH} = 0.0150\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.00300\text{ mol}\). Moles of \(\text{HCl}\) reacted with \(\text{CaCO}_3 = 0.0250 - 0.00300 = 0.0220\text{ mol}\). Since 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of \(\text{HCl}\), moles of \(\text{CaCO}_3 = 0.0220 / 2 = 0.0110\text{ mol}\). Mass of \(\text{CaCO}_3 = 0.0110\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.1011\text{ g}\). Percentage by mass = \((1.1011\text{ g} / 1.50\text{ g}) \times 100\% \approx 73.4\%\).

1 mark for the correct answer (A). 0 marks for incorrect options.

Both methyl propanoate and ethyl ethanoate are isomeric esters. They contain exactly the same types of bonds: C=O, C-O, and C-H, resulting in identical major functional group absorption peaks in their IR spectra. Thus, they cannot be distinguished by identifying characteristic functional group absorptions alone. The other pairs have different functional groups and can be easily distinguished by identifying the presence or absence of specific peaks (e.g., O-H vs C=O).

1 mark for the correct answer (D). 0 marks for incorrect options.

\(\text{BF}_3\) has 3 bonding pairs and 0 lone pairs on B, which is trigonal planar. \(\text{CO}_3^{2-}\) and \(\text{NO}_3^-\) both have 3 bonding regions and 0 lone pairs on the central carbon and nitrogen atoms respectively, which also gives them a trigonal planar shape. In B, the species are trigonal pyramidal. In C, they are tetrahedral. In D, they are bent.

1 mark for the correct answer (A). 0 marks for incorrect options.

Atom economy = (molar mass of desired product / sum of molar masses of all products) x 100%. Desired product is maleic anhydride (\(\text{C}_4\text{H}_2\text{O}_3\)), molar mass = \(4 \times 12.0 + 2 \times 1.0 + 3 \times 16.0 = 98.0\text{ g mol}^{-1}\). Byproducts are 3 moles of \(\text{H}_2\text{O}\), mass = \(3 \times 18.0 = 54.0\text{ g}\). Sum of molar masses of all products = \(98.0 + 54.0 = 152.0\text{ g}\). Atom economy = \((98.0 / 152.0) \times 100\% = 64.5\%\).

1 mark for the correct answer (A). 0 marks for incorrect options.

According to Hess's Law, the equation for the combustion of liquid ethanol is:
\( \text{C}_2\text{H}_5\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)} \)
Using the standard enthalpy of formation (\( \Delta H_f^{\theta} \)) of the substances:
\( \Delta H_c^{\theta} = 2 \Delta H_1 + 3 \Delta H_2 - \Delta H_3 \)
\( \Delta H_c^{\theta} = 2(-393.5) + 3(-285.8) - (-277.6) = -1366.8 \text{ kJ mol}^{-1} \).

Award 1 mark for the correct option A. Method involves applying Hess's Law: \( 2 \times \Delta H_1 + 3 \times \Delta H_2 - \Delta H_3 \).

In option A, elemental sulfur (oxidation number = 0) is oxidized to sulfur dioxide (\( \text{SO}_2 \), oxidation number of S = +4). The increase is indeed 4. In C, sulfur dioxide (+4) is oxidized to sulfate (+6), change is +2. In D, sulfuric acid (+6) is reduced to sulfur dioxide (+4), change is -2.

Award 1 mark for selecting A. Check the initial and final oxidation states of S in each option.

1. Longest carbon chain containing the double bond is 5 carbons (pentene).
2. Numbering from right to left gives the double bond the lowest locant: C2 (pent-2-ene).
3. Substituents are methyl groups at carbon-3 and carbon-4. Thus, the name is 3,4-dimethylpent-2-ene.

Award 1 mark for B. Locant for double bond has priority over substituents for lower numbers.

1. Aldehydes (option A) are oxidized by acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \).
2. Alcohols (options B and D) react with sodium metal to release hydrogen gas and can be oxidized.
3. Ketones (option C, butanone) have molecular formula \( \text{C}_4\text{H}_8\text{O} \), are resistant to mild oxidation, and do not react with sodium metal.

Award 1 mark for C. Students need to deduce functional groups based on the chemical tests.

According to the Arrhenius equation, \( \ln k = -\frac{E_a}{R} (\frac{1}{T}) + \ln A \). The slope \( m = -\frac{E_a}{R} \). Thus, \( -1.20 \times 10^4 = -\frac{E_a}{8.31} \) which gives \( E_a = 99720 \text{ J mol}^{-1} \approx 99.7 \text{ kJ mol}^{-1} \).

Award 1 mark for C. Proper unit conversion (J to kJ) is required.

Formula mass of \( \text{C}_4\text{H}_2\text{O}_3 = 98.0 \).
For Route 1: Desired product mass = \( 2 \times 98.0 = 196.0 \). Total mass of products = \( 2(98.0) + 4(44.0) + 4(18.0) = 444.0 \). Atom economy = \( \frac{196.0}{444.0} \times 100\% = 44.1\% \).
For Route 2: Desired product mass = \( 98.0 \). Total mass of products = \( 98.0 + 3(18.0) = 152.0 \). Atom economy = \( \frac{98.0}{152.0} \times 100\% = 64.5\% \).
Route 2 is greener due to its higher atom economy (64.5%).

Award 1 mark for correct selection B with correct calculations.

1. Moles of \( \text{S}_2\text{O}_3^{2-} \) used = \( 0.100 \times 0.02240 = 2.24 \times 10^{-3} \text{ mol} \).
2. Moles of \( \text{I}_2 = \frac{1}{2} \times n(\text{S}_2\text{O}_3^{2-}) = 1.12 \times 10^{-3} \text{ mol} \).
3. Moles of \( \text{OCl}^- = n(\text{I}_2) = 1.12 \times 10^{-3} \text{ mol} \) in \( 25.00 \text{ cm}^3 \) diluted solution.
4. Conc. of \( \text{OCl}^- \) in diluted solution = \( 1.12 \times 10^{-3} / 0.02500 = 0.0448 \text{ mol dm}^{-3} \).
5. Original conc. (dilution factor of 10) = \( 0.0448 \times 10 = 0.448 \text{ mol dm}^{-3} \).

Award 1 mark for option C. Method involves sequential stoichiometric ratios and accounting for the dilution factor.

From the IR data, the compound contains an O-H group (carboxylic acid, broad at \( 3000 \text{ cm}^{-1} \)) and a C=O group (sharp at \( 1715 \text{ cm}^{-1} \)). Therefore, \( Y \) is propanoic acid (\( \text{CH}_3\text{CH}_2\text{COOH} \)). In the mass spectrum of propanoic acid, cleavage adjacent to the carbonyl group yields the highly stable acylium/carboxyl fragment \( [\text{COOH}]^+ \) at \( m/z = 45 \).

Award 1 mark for B. Student must identify the compound as propanoic acid first, then identify the correct mass fragment.

According to the metal reactivity series: \( \text{Al} > \text{Zn} > \text{Fe} > \text{Pb} > \text{Cu} \). A free metal can displace a less reactive metal from its oxide. Since aluminum (\( \text{Al} \)) is more reactive than iron (\( \text{Fe} \)), heating a mixture of \( \text{Fe}_2\text{O}_3 \) and \( \text{Al} \) results in a vigorous thermite reaction. Fe, Cu, and Pb are less reactive than Mg, Al, and Zn respectively, so other options show no reaction.

Award 1 mark for C. The question tests understanding of the reactivity series applied to solid-state displacement reactions.

\( \text{H}_3\text{O}^+ \) has 3 bond pairs and 1 lone pair on the central oxygen atom, resulting in a trigonal pyramidal shape. \( \text{PCl}_3 \) also has 3 bond pairs and 1 lone pair on the central phosphorus atom, which also gives it a trigonal pyramidal shape. \( \text{BF}_3 \) and \( \text{CO}_3^{2-} \) are trigonal planar, while \( \text{ClF}_3 \) is T-shaped.

Award 1 mark for B. Apply VSEPR theory to find electron pairs and shapes of molecules.

According to Hess's Law, the equation for the formation of propane is: \(3\text{C(graphite)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\). Using the standard enthalpy changes of combustion: \(\Delta H_f^\theta = 3 \times \Delta H_c^\theta[\text{C}] + 4 \times \Delta H_c^\theta[\text{H}_2] - \Delta H_c^\theta[\text{C}_3\text{H}_8]\) which gives \(\Delta H_f^\theta = 3(-393.5) + 4(-285.8) - (-2220.0) = -103.7\text{ kJ mol}^{-1}\).

Award 1 mark for correct choice A. Method: apply Hess's Law to relate standard enthalpies of combustion to standard enthalpy of formation. Accuracy: calculation gives -103.7 kJ/mol.

(1) is correct because the original cell voltage is 0.34 - (-0.76) = 1.10 V, while the new cell voltage is 0.34 - (-0.25) = 0.59 V. (2) is correct because Ni is more reactive than Cu, so Ni acts as the negative electrode and undergoes oxidation, releasing electrons which flow to the Cu electrode. (3) is incorrect because Ni undergoes oxidation to Ni2+(aq), so the nickel electrode dissolves and its mass decreases.

Award 1 mark for correct choice B. Deduce that (1) and (2) are correct, while (3) is incorrect because nickel dissolves during discharge.

The longest carbon chain containing the double bond has 5 carbon atoms, making it a pentene. Numbering from right to left gives the double bond the lowest locants (at C2). The substituent at C1 is chlorine, and the substituent at C4 is methyl. Thus, the IUPAC name is 1-chloro-4-methylpent-2-ene.

Award 1 mark for correct choice B. Correctly identify the parent chain length (pent-2-ene) and assign lowest locants priority to the double bond over the alkyl/halogen groups.

Compound Z is an ester with 5 carbon atoms. Since it is formed by esterification of Compound Y with ethanol (which has 2 carbon atoms), Compound Y must be a carboxylic acid with 3 carbon atoms (propanoic acid). Compound Y is formed by oxidation under reflux of Compound X. Thus, X must be a primary alcohol with 3 carbon atoms, which is propan-1-ol.

Award 1 mark for correct choice A. Deduce the ester structure as ethyl propanoate, identifying Y as propanoic acid, and X as propan-1-ol.

Initial moles of HCl = 0.0500 * 0.500 = 0.0250 mol. Moles of NaOH used = 0.0168 * 0.250 = 0.0042 mol. Moles of HCl in excess = 0.0042 mol. Moles of HCl reacted with CaCO3 = 0.0250 - 0.0042 = 0.0208 mol. Since CaCO3 + 2HCl -> CaCl2 + CO2 + H2O, moles of CaCO3 = 0.0208 / 2 = 0.0104 mol. Mass of CaCO3 = 0.0104 * 100.1 = 1.041 g. Mass percentage = (1.041 / 2.50) * 100% = 41.6%.

Award 1 mark for correct choice B. Method: calculate excess moles of HCl, find moles of HCl reacted, use 1:2 ratio to find moles of CaCO3, and calculate mass percentage.

According to the Arrhenius equation: \(\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A\). The slope of the plot is \(-\frac{E_a}{R}\). Therefore, \(-\frac{E_a}{R} = -1.20 \times 10^4\text{ K}\), giving \(E_a = 1.20 \times 10^4 \times 8.314\text{ J mol}^{-1} = 9.98 \times 10^4\text{ J mol}^{-1} = 99.8\text{ kJ mol}^{-1}\).

Award 1 mark for correct choice A. Method: relate slope to -Ea/R and perform conversion from Joules to kilojoules.

Formula mass of Aniline = 6(12.0) + 7(1.0) + 14.0 = 93.0. Formula mass of Acetic anhydride = 4(12.0) + 6(1.0) + 3(16.0) = 102.0. Total mass of reactants = 93.0 + 102.0 = 195.0. Formula mass of desired product Acetanilide = 8(12.0) + 9(1.0) + 14.0 + 16.0 = 135.0. Atom economy = (135.0 / 195.0) * 100% = 69.2%.

Award 1 mark for correct choice C. Method: calculate the ratio of the molar mass of the desired product to the sum of the molar masses of all reactants, expressed as a percentage.

The infrared absorption band at 3230 - 3670 cm-1 indicates the presence of an O-H group (alcohol). Primary alcohols (propan-1-ol) and secondary alcohols (propan-2-ol) are oxidized by acidified potassium dichromate, changing the solution's color from orange to green. Propanoic acid contains a carboxyl group (O-H stretch at 2500 - 3300 cm-1). 2-methylpropan-2-ol is a tertiary alcohol, which cannot be oxidized by acidified dichromate, hence causing no color change.

Award 1 mark for correct choice C. Method: identify the infrared band as an alcohol O-H stretch and determine which alcohol is resistant to dichromate oxidation.

Observation (1) indicates P is highly reactive (e.g., alkali/alkaline earth metal). Observation (2) shows P is more reactive than Q because the oxide of P is too stable to be reduced by carbon, unlike the oxide of Q. Observation (3) shows Q is more reactive than R because Q displaces R from solution. Therefore, the decreasing order of reactivity is P > Q > R.

Award 1 mark for correct choice A. Deduce reactivity based on reaction with cold water, ease of oxide reduction by carbon, and metal displacement.

Both CH4 and NH4+ have a central atom bonded to 4 terminal atoms with 0 lone pairs, giving them a tetrahedral shape. BF3 (trigonal planar) and NH3 (trigonal pyramidal) have different shapes. H2O (bent) and CO2 (linear) have different shapes. PCl3 (trigonal pyramidal) and BF3 (trigonal planar) have different shapes.

Award 1 mark for correct choice B. Apply VSEPR theory to identify that both CH4 and NH4+ possess a tetrahedral shape.

By definition, the standard enthalpy change of formation of ethyne corresponds to the reaction:
\( 2\text{C(graphite)} + \text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_2\text{(g)} \)

Using Hess's Law and standard enthalpies of combustion:
For the combustion of ethyne:
\( \text{C}_2\text{H}_2\text{(g)} + 2.5\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + \text{H}_2\text{O(l)} \)
\( \Delta H_c^\theta[\text{C}_2\text{H}_2\text{(g)}] = \frac{-2598.8}{2} = -1299.4 \text{ kJ mol}^{-1} \)

The standard enthalpy of combustion of graphite is:
\( \Delta H_c^\theta[\text{C(graphite)}] = -393.5 \text{ kJ mol}^{-1} \)

The standard enthalpy of combustion of hydrogen is:
\( \Delta H_c^\theta[\text{H}_2\text{(g)}] = \frac{-571.6}{2} = -285.8 \text{ kJ mol}^{-1} \) (which is also the enthalpy of formation of \(\text{H}_2\text{O(l)}\))

Using Hess's Law:
\( \Delta H_c^\theta[\text{C}_2\text{H}_2\text{(g)}] = 2\Delta H_f^\theta[\text{CO}_2\text{(g)}] + \Delta H_f^\theta[\text{H}_2\text{O(l)}] - \Delta H_f^\theta[\text{C}_2\text{H}_2\text{(g)}] \)
\( -1299.4 = 2(-393.5) + (-285.8) - \Delta H_f^\theta[\text{C}_2\text{H}_2\text{(g)}] \)
\( -1299.4 = -787.0 - 285.8 - \Delta H_f^\theta[\text{C}_2\text{H}_2\text{(g)}] \)
\( -1299.4 = -1072.8 - \Delta H_f^\theta[\text{C}_2\text{H}_2\text{(g)}] \)
\( \Delta H_f^\theta[\text{C}_2\text{H}_2\text{(g)}] = -1072.8 - (-1299.4) = +226.6 \text{ kJ mol}^{-1} \).

Award 1 mark for the correct option C. No partial marks.

Let \(x\) be the oxidation number of vanadium in each species.
- For (1) \(\text{VO}^{2+}\): \(x + (-2) = +2 \Rightarrow x = +4\).
- For (2) \(\text{VO}_2\): \(x + 2(-2) = 0 \Rightarrow x = +4\).
- For (3) \(\text{VCl}_4\): \(x + 4(-1) = 0 \Rightarrow x = +4\).
Therefore, all three species contain vanadium with an oxidation state of +4.

Award 1 mark for the correct option D. No partial marks.

With the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\), the compound has one degree of unsaturation (a ring or a double bond).
- It does not react with \(\text{Na}\) metal, meaning it does not contain an \(-\text{OH}\) group (excluding alcohols, diols, and carboxylic acids).
- It does not react with \(\text{NaHCO}_3\) solution, meaning it is not a carboxylic acid.
- Since it contains two oxygen atoms, lacks \(-\text{OH}\) and \(-\text{COOH}\) groups, and has one double bond (the \(\text{C}=\text{O}\) group), it is most likely an ester (e.g., ethyl ethanoate).

Award 1 mark for the correct option B. No partial marks.

- (1) Increasing temperature increases the kinetic energy of the molecules. The Maxwell-Boltzmann distribution curve shifts to the right, increasing the fraction of molecules with \(E \ge E_a\), thus increasing the reaction rate.
- (2) Adding a catalyst provides an alternative pathway with a lower activation energy (\(E_a'\)). Because \(E_a' < E_a\), the fraction of molecules with energy equal to or greater than the active activation energy increases, which increases the reaction rate.
- (3) Increasing pressure increases the concentration of reactants and collision frequency, which increases the reaction rate, but it does NOT change the kinetic energy of the molecules or the activation energy. Thus, the fraction of reactant molecules with energy \(\ge E_a\) remains unchanged.
Therefore, only (1) and (2) are correct.

Award 1 mark for the correct option B. No partial marks.

The carbon skeleton of the acyclic compound must be a propane chain (\(\text{C}-\text{C}-\text{C}\)). We can arrange the three chlorine atoms on this chain as follows:
1. 1,1,1-trichloropropane: \(\text{CCl}_3-\text{CH}_2-\text{CH}_3\)
2. 1,1,2-trichloropropane: \(\text{CHCl}_2-\text{CHCl}-\text{CH}_3\)
3. 1,1,3-trichloropropane: \(\text{CHCl}_2-\text{CH}_2-\text{CH}_2\text{Cl}\)
4. 1,2,2-trichloropropane: \(\text{CH}_2\text{Cl}-\text{CCl}_2-\text{CH}_3\)
5. 1,2,3-trichloropropane: \(\text{CH}_2\text{Cl}-\text{CHCl}-\text{CH}_2\text{Cl}\)
Note that the central carbon can hold at most two chlorine atoms. All other positions are equivalent to these 5 by symmetry. Thus, there are exactly 5 structural isomers.

Award 1 mark for the correct option B. No partial marks.

The calculation formula is: \(C_{\text{HCl}} = \frac{2 \times C_{\text{Na}_2\text{CO}_3} \times V_{\text{Na}_2\text{CO}_3}}{V_{\text{HCl}}}\).
- A: Rinsing the pipette with distilled water only dilutes the sodium carbonate solution. The actual number of moles of \(\text{Na}_2\text{CO}_3\) transferred is less than calculated. Consequently, a smaller volume of \(\text{HCl}\) (\(V_{\text{HCl}}\)) is needed to reach the end point. Since a smaller \(V_{\text{HCl}}\) is recorded, the calculated concentration of \(\text{HCl}\) will be higher than the actual value (overestimation). (Correct!)
- B: Rinsing the conical flask with sodium carbonate solution leaves extra \(\text{Na}_2\text{CO}_3\) in the flask, meaning a larger volume of \(\text{HCl}\) is required, resulting in an underestimation.
- C: Rinsing the burette with distilled water only dilutes the \(\text{HCl}\), meaning a larger volume of this diluted acid is needed, resulting in an underestimation.
- D: Dislodging an air bubble during titration increases the recorded volume of \(\text{HCl}\) delivered, resulting in an underestimation.

Award 1 mark for the correct option A. No partial marks.

(a) The anode should be made of the impure copper sample because oxidation occurs at the anode, where copper atoms are oxidised to copper(II) ions. (b) Zinc is more reactive than copper, so it is oxidised to zinc(II) ions and enters the electrolyte. Silver is less reactive than copper, so it is not oxidised and falls to the bottom of the container as anode slime. (c) The ionic half-equation at the cathode is Cu2+(aq) + 2e- -> Cu(s). A reddish-brown solid is deposited on the cathode.

(a) Anode (1 mark); Copper is oxidised to Cu2+ (1 mark). (b) Zinc oxidised to Zn2+ in solution (1 mark); Silver remains unreacted as anode slime (1 mark). (c) Cu2+(aq) + 2e- -> Cu(s) (1 mark); Reddish-brown solid deposited (1 mark).

(a) The systematic IUPAC name of monomer X is ethyl 2-methylpropenoate. (b) The repeating unit is -[CH2-C(CH3)(COOCH2CH3)]-. (c) The polymer consists of long polymer chains held together by weak intermolecular forces (Van der Waals' forces) without covalent cross-links between them. Upon heating, these weak forces are easily overcome, allowing the chains to slide over one another. (d) Incineration of the carbon-based polymer releases carbon dioxide (a greenhouse gas) which contributes to global warming, or carbon monoxide due to incomplete combustion.

(a) Ethyl 2-methylpropenoate (1 mark). (b) Correct repeating unit structure with open single bonds on both ends (1 mark). (c) Held by weak intermolecular forces / no covalent cross-links (1 mark); Chains can slide past each other when heated (1 mark). (d) CO2 gas released leading to enhanced greenhouse effect / toxic CO gas from incomplete combustion (2 marks).

(a) Titanium is highly reactive and has a high affinity for carbon; heating titanium(IV) oxide with carbon would produce titanium carbide (TiC) instead of pure titanium metal. (b) The chemical equation is TiCl4 + 2Mg -> Ti + 2MgCl2. (c) An inert atmosphere of argon is used to prevent the hot magnesium and titanium from reacting with oxygen and nitrogen in the air. (d) Advantage: Recycling conserves limited titanium ore resources and saves a significant amount of energy. Disadvantage: Sorting and purifying titanium scrap is difficult and expensive.

(a) Titanium is highly reactive / has high affinity for carbon (1 mark); Heating with carbon forms titanium carbide (1 mark). (b) TiCl4 + 2Mg -> Ti + 2MgCl2 (1 mark). (c) Prevents hot Mg/Ti from reacting with oxygen/nitrogen in air (1 mark). (d) One advantage: saves energy / conserves resource (1 mark); One disadvantage: high cost of collection/sorting/purifying (1 mark).

(a) The thermochemical equation is: 3C(s, graphite) + 4H2(g) + 0.5O2(g) -> CH3CH(OH)CH3(l) \Delta H_f^\theta = -320 kJ mol-1. (b) According to Hess's Law, the enthalpy change of the reaction can be calculated by comparing the combustion products of both sides. Enthalpy of reactants combustion: 3 * (-394) + 4 * (-286) = -1182 - 1144 = -2326 kJ mol-1. Enthalpy of product combustion: -2006 kJ mol-1. \Delta H_f^\theta = \sum \Delta H_c^\theta(reactants) - \sum \Delta H_c^\theta(products) = -2326 - (-2006) = -320 kJ mol-1.

(a) Correct equation with balanced coefficients (1 mark); Correct state symbols (1 mark). (b) Drawing a correct Hess's Law cycle or expressing correct algebraic relationship (1 mark); Calculating total reactant combustion enthalpy as -2326 kJ mol-1 (1 mark); Expressing \Delta H_f^\theta = -2326 - (-2006) (1 mark); Final answer: -320 kJ mol-1 with unit (1 mark).

(a) The catalyst is concentrated sulfuric acid. Its other function is acting as a dehydrating agent (or removing water to shift the equilibrium position to the right). (b) Heating under reflux prevents the loss of volatile organic reactants (ethanol and ethanoic acid) and product (ethyl ethanoate) by condensing their vapours back into the reaction flask, allowing the reaction to be heated for a longer duration. (c) Sodium carbonate is added to neutralise excess unreacted ethanoic acid and the sulfuric acid catalyst. Completion is indicated when no more gas bubbles (effervescence) are produced.

(a) Concentrated sulfuric acid (1 mark); Dehydrating agent / shifts equilibrium (1 mark). (b) Prevents loss of volatile reactants/products (1 mark); Condenses vapour back to allow heating for longer time (1 mark). (c) Neutralise excess unreacted acids (1 mark); Effervescence stops / no more gas bubbles (1 mark).

(a) Using the formula: ln(6.8 * 10^-2 / 2.4 * 10^-3) = -Ea / 8.31 * (1/340 - 1/300). ln(28.333) = -Ea / 8.31 * (-0.000392). 3.344 = Ea / 8.31 * 0.000392. Ea = 3.344 * 8.31 / 0.000392 = 70847 J mol-1 = 70.8 kJ mol-1. (b) A catalyst provides an alternative reaction pathway with a lower activation energy. Therefore, a larger fraction of reactant molecules possess kinetic energy greater than or equal to the activation energy, which increases the frequency of effective collisions.

(a) Correct substitution of k and T values into Arrhenius equation (1 mark); Evaluation of ln(k2/k1) and temperature term (1 mark); Accurate value of Ea in Joules (70847 J mol-1) (1 mark); Final value converted to kJ mol-1 (70.8 kJ mol-1) (1 mark). (b) Catalyst provides alternative pathway with lower Ea (1 mark); Larger fraction of reactant molecules have kinetic energy >= Ea to increase effective collision frequency (1 mark).

(a) Atom economy of Route A = (molar mass of P) / (molar mass of X + Y) * 100% = 120 / (80 + 90) * 100% = 120 / 170 * 100% = 70.6%. (b) The starting reactants for the overall Route B are A, B, and D (as C is an intermediate). Atom economy of Route B = (molar mass of P) / (molar mass of A + B + D) * 100% = 120 / (50 + 70 + 10) * 100% = 120 / 130 * 100% = 92.3%. (c) 1. Route B has a higher atom economy, which reduces chemical waste. 2. Route B avoids the production of by-product Z, which might be hazardous or require additional cost to dispose of.

(a) Calculation setup (120/170) (1 mark); Correct percentage 70.6% (1 mark). (b) Identifying reactants as A, B, D and setup (120/130) (1 mark); Correct percentage 92.3% (1 mark). (c) Route B has higher atom economy/less waste (1 mark); Avoids production of hazardous by-product Z (1 mark).

(a) CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g). (b) Initial moles of HCl = 1.00 * (50.0/1000) = 0.0500 mol. Moles of NaOH used in titration = 0.100 * (14.10/1000) = 0.00141 mol. Moles of HCl in 25.0 cm3 filtrate = 0.00141 mol (since reaction ratio of HCl to NaOH is 1:1). Total moles of excess HCl in 250.0 cm3 filtrate = 0.00141 * (250.0/25.0) = 0.0141 mol. Moles of HCl reacted with CaCO3 = 0.0500 - 0.0141 = 0.0359 mol. (c) From the equation, 1 mole of CaCO3 reacts with 2 moles of HCl. Moles of CaCO3 = 0.0359 / 2 = 0.01795 mol. Mass of CaCO3 = 0.01795 * 100.1 = 1.797 g. Percentage by mass = (1.797 / 2.00) * 100% = 89.8%.

(a) CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g) (1 mark). (b) Initial moles of HCl = 0.0500 mol (1 mark); Moles of HCl in 25.0 cm3 = 0.00141 mol (1 mark); Total excess moles of HCl = 0.0141 mol (1 mark); Moles of reacted HCl = 0.0359 mol (1 mark). (c) Percentage by mass = 89.8% (1 mark).

(a) The carbonyl isomers of C3H6O are propanal (CH3CH2CHO) and propanone (CH3COCH3). (b) The strong peak at 1715 cm-1 indicates the presence of a C=O (carbonyl) group. Since there is no absorption in the region of 2700 - 2830 cm-1 (which is characteristic of the aldehydic C-H stretching vibration), W does not contain an aldehyde group. Hence, W is a ketone. (c) Warm both compounds separately with Tollens' reagent (ammoniacal silver nitrate). Propanal (the aldehyde) will form a silver mirror on the inner wall of the test tube, while propanone (the ketone) will show no observable change.

(a) Structures of propanal and propanone (1 mark for each). (b) Carbonyl group / C=O (1 mark); Ketone, as it lacks the aldehyde C-H absorption at 2700-2830 cm-1 (1 mark). (c) Tollens' reagent / Fehling's solution (1 mark); Aldehyde gives silver mirror / red ppt, ketone has no change (1 mark).

(a) Kc = [CO][Cl2] / [COCl2]. (b) Initial concentration of COCl2 = 1.00 / 2.00 = 0.50 mol dm-3. At equilibrium, [CO] = 0.15 mol dm-3, so [Cl2] = 0.15 mol dm-3 because of the 1:1 mole ratio. [COCl2] at equilibrium = 0.50 - 0.15 = 0.35 mol dm-3. Substituting these values into the Kc expression: Kc = (0.15 * 0.15) / 0.35 = 0.0643 mol dm-3. (c) Since the forward reaction is endothermic (\Delta H > 0), according to Le Chatelier's principle, increasing the temperature will shift the equilibrium position to the right (the endothermic direction) to absorb the added thermal energy. Thus, the equilibrium yield of CO(g) will increase.

(a) Kc = [CO][Cl2]/[COCl2] (1 mark). (b) Equilibrium [Cl2] = 0.15 mol dm-3 and [COCl2] = 0.35 mol dm-3 (1 mark); Correct Kc calculation setup (1 mark); Final answer: 0.0643 mol dm-3 (accept 0.064) with correct unit (1 mark). (c) Yield of CO increases (1 mark); Forward reaction is endothermic, so equilibrium shifts to the right to absorb heat (1 mark).

(a) At the anode (oxidation): \(\text{H}_2(g) + 2\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + 2e^-\)
(b) At the cathode (reduction): \(\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq)\)
(c) Fuel cells have a higher energy conversion efficiency than conventional dry cells, or they can operate continuously as long as external fuel is supplied (no need for disposal/recharging).
(d)
1. Charge passed: \(Q = I \times t = 2.0\text{ A} \times 9650\text{ s} = 19300\text{ C}\).
2. Moles of electrons: \(n(e^-) = \frac{19300\text{ C}}{96500\text{ C mol}^{-1}} = 0.20\text{ mol}\).
3. From the half-equations, 2 moles of electrons correspond to 1 mole of \(\text{H}_2\text{O}\) produced overall (or anode half-reaction ratio: \(2e^- : 2\text{H}_2\text{O}\), but with cathode consumption, overall stoichiometry is \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\) where 4 electrons are transferred for every 2 molecules of water formed, i.e., \(2e^-\) per \(\text{H}_2\text{O}\)).
Hence, \(n(\text{H}_2\text{O}) = \frac{1}{2} n(e^-) = 0.10\text{ mol}\).
4. Mass of \(\text{H}_2\text{O} = 0.10\text{ mol} \times 18.0\text{ g mol}^{-1} = 1.80\text{ g}\).

(a) [1 mark] for \(\text{H}_2(g) + 2\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + 2e^-\). Do not accept acidic half-equations.
(b) [1 mark] for \(\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightarrow 4\text{OH}^-(aq)\). Do not accept acidic half-equations.
(c) [1 mark] for stating high efficiency, continuous fuel supply, or non-polluting product (water).
(d) [1 mark] for calculating charge and moles of electrons (0.20 mol); [1 mark] for finding the correct mole ratio of electron to water (2:1, hence 0.10 mol of water); [1 mark] for calculating the correct mass of water (1.80 g). Deduct 1 mark for incorrect/missing unit.

(a) The standard enthalpy change of combustion is the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions (1 atm, 298 K).
(b) The elements in their standard states are carbon as graphite (solid) and hydrogen as diatomic gas: \(3\text{C(graphite, } s) + 3\text{H}_2(g) \rightarrow \text{C}_3\text{H}_6(g)\).
(c) According to Hess's Law:
\(\Delta H_f^⊖[\text{C}_3\text{H}_6(g)] = 3 \times \Delta H_c^⊖[\text{C(graphite)}] + 3 \times \Delta H_c^⊖[\text{H}_2(g)] - \Delta H_c^⊖[\text{C}_3\text{H}_6(g)]\)
\(\Delta H_f^⊖[\text{C}_3\text{H}_6(g)] = 3(-394) + 3(-286) - (-2058)\)
\(\Delta H_f^⊖[\text{C}_3\text{H}_6(g)] = -1182 - 858 + 2058 = +18\text{ kJ mol}^{-1}\).

(a) [1 mark] for "one mole of substance"; [1 mark] for "completely burned in oxygen under standard conditions".
(b) [1 mark] for the correct balanced equation with state symbols (\(3\text{C(graphite, s)} + 3\text{H}_2(g) \rightarrow \text{C}_3\text{H}_6(g)\)). Accept C(s).
(c) [1 mark] for showing a correct enthalpy cycle or mathematical expression based on Hess's Law; [1 mark] for substituting correct numbers into the expression; [1 mark] for calculating the correct answer with unit (\(+18\text{ kJ mol}^{-1}\) or \(18\text{ kJ mol}^{-1}\)).

(a) Reaction 1 is nucleophilic substitution. The halogenoalkane is converted to alcohol using aqueous sodium hydroxide (\(\text{NaOH}(aq)\)) or aqueous potassium hydroxide (\(\text{KOH}(aq)\)) under heating / reflux conditions.
(b) Reaction 2 is oxidation of primary alcohol to carboxylic acid. Reagent B is acidified potassium dichromate solution (\(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+(aq)\)), which oxidizes propan-1-ol to propanoic acid. The orange dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) is reduced to green chromium(III) ion (\(\text{Cr}^{3+}\)), giving a color change from orange to green.
(c) Reaction 3 is esterification. Reagent C is concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)), which acts as a catalyst to speed up the reaction and as a dehydrating agent to shift the equilibrium position to the right.

(a) [1 mark] for \(\text{NaOH}(aq)\) or \(\text{KOH}(aq)\) (reject NaOH without state, reject dilute sulfuric acid); [1 mark] for heating under reflux.
(b) [1 mark] for acidified potassium dichromate solution / \(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+(aq)\) (accept acidified potassium permanganate / \(\text{KMnO}_4 / \text{H}^+(aq)\)); [1 mark] for orange to green (or purple to colorless if permanganate is used).
(c) [1 mark] for concentrated sulfuric acid / conc. \(\text{H}_2\text{SO}_4\) (reject sulfuric acid without "concentrated"); [1 mark] for catalyst / dehydrating agent.

(a) (i)
Using the Arrhenius equation:
\(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
Substitute the values into the equation:
\(\ln\left(\frac{1.5 \times 10^{-2}}{2.5 \times 10^{-3}}\right) = \frac{E_a}{8.31}\left(\frac{1}{500} - \frac{1}{550}\right)\)
\(\ln(6) = \frac{E_a}{8.31}\left(2.00 \times 10^{-3} - 1.818 \times 10^{-3}\right)\)
\(1.7918 = \frac{E_a}{8.31}\left(1.818 \times 10^{-4}\right)\)
\(E_a = \frac{1.7918 \times 8.31}{1.818 \times 10^{-4}} = 81892\text{ J mol}^{-1}\)
\(E_a = 81.9\text{ kJ mol}^{-1}\) (or \(82\text{ kJ mol}^{-1}\))

(ii)
- Sketch:
- Axes correctly labeled: Y-axis as "Fraction of molecules" (or "Number of molecules with energy E"), X-axis as "Kinetic energy".
- The curve for \(550\text{ K}\) is flattened and shifted to the right compared to the curve for \(500\text{ K}\).
- The activation energy \(E_a\) is indicated on the X-axis, showing a larger shaded area (fraction of molecules) under the curve at \(550\text{ K}\) than at \(500\text{ K}\).
- Explanation: At a higher temperature (\(550\text{ K}\)), the average kinetic energy of the molecules increases. Thus, a larger fraction of molecules have kinetic energy equal to or greater than the activation energy (\(E \ge E_a\)), leading to an increased frequency of effective collisions, and hence a faster reaction rate.

(b) (i)
- Temperature: Since the forward reaction is exothermic (\(\Delta H < 0\)), increasing the temperature shifts the equilibrium position to the left (reactant side) to absorb the added heat. Thus, the equilibrium yield of methanol decreases.
- Pressure: There are 4 moles of gas reactants (\(1\text{ CO}_2 + 3\text{ H}_2\)) and 2 moles of gas products (\(1\text{ CH}_3\text{OH} + 1\text{ H}_2\text{O}\)). Increasing pressure shifts the equilibrium position to the right (product side) to reduce the total number of gas molecules and relieve the pressure. Thus, the equilibrium yield of methanol increases.

(ii)
- Kinetic factor: High temperature is needed to obtain a high rate of reaction so that equilibrium is reached quickly.
- Thermodynamic factor: Low temperature is favored for high equilibrium yield of methanol (as the forward reaction is exothermic). Hence, \(250^\circ\text{C}\) is a compromise temperature between reaction rate and yield.
- Economic/Safety factor: High pressure increases both reaction rate and equilibrium yield, but extremely high pressures require expensive thick-walled reactors, high compression energy costs, and pose safety hazards. \(50\text{ atm}\) is a compromise pressure balancing yield against equipment costs and safety.

(c) (i)
Molar mass of desired product (\(CH_3OH\)) = \(12.0 + (1.0 \times 4) + 16.0 = 32.0\text{ g mol}^{-1}\)
- Pathway 1:
Total mass of reactants = \(M(CO) + 2 \times M(H_2) = (12.0 + 16.0) + 2 \times (2.0) = 32.0\text{ g mol}^{-1}\)
\(\text{Atom economy} = \frac{32.0}{32.0} \times 100\% = 100\%\)
- Pathway 2:
Total mass of reactants = \(M(CO_2) + 3 \times M(H_2) = (12.0 + 16.0 \times 2) + 3 \times (2.0) = 50.0\text{ g mol}^{-1}\)
\(\text{Atom economy} = \frac{32.0}{50.0} \times 100\% = 64.0\%\)

(ii)
- Pathway 2 utilizes carbon dioxide (\(\text{CO}_2\)), which is a major greenhouse gas and industrial waste, thus contributing to carbon capture and reducing global warming.
- In contrast, \(\text{CO}\) in Pathway 1 is highly toxic and typically derived from non-renewable fossil fuels. Additionally, if the hydrogen gas (\(\text{H}_2\)) is obtained from water electrolysis powered by renewable energy, Pathway 2 enables a carbon-neutral cycle.

(iii) (1)
A catalyst increases the reaction rate by providing an alternative reaction pathway with a lower activation energy.

(iii) (2)
- Catalyst poisoning: Sulfur compounds (e.g., \(\text{H}_2\text{S}\)) adsorb strongly and irreversibly onto the active sites on the copper catalyst surface, blocking the reactants from accessing these active sites.
- Prevention: Pre-treat the feed gases by passing them through a desulfurization bed (such as a zinc oxide guard bed) to remove sulfur impurities before they reach the catalyst.

(a) (i)
- Correct Arrhenius equation formula (1)
- Correct substitution of values into formula (1)
- Correct calculation of E_a in Joules (81892 J mol^-1) (1)
- Correct conversion to kJ mol^-1 with appropriate decimal/sig figs and unit (81.9 or 82 kJ mol^-1) (1)

(ii)
- Correctly labeled axes AND correct shifting of 550 K curve (lower peak, shifted to the right) with E_a marked (2)
- Explanation based on collision theory: higher temperature increases fraction of molecules with E >= Ea, leading to increased frequency of effective collisions (1)

(b) (i)
- Predicts and explains decrease in yield with temperature based on Le Chatelier's Principle / exothermic nature (2)
- Predicts and explains increase in yield with pressure based on change in number of gaseous moles (2)

(ii)
- Explains 250 °C as a compromise between rate (favored by high T) and yield (favored by low T) (2)
- Explains 50 atm as a compromise between rate/yield (favored by high P) and economic/safety costs (1)

(c) (i)
- Correct calculation of molar mass of desired product and reactants (1)
- Correct atom economy for Pathway 1 (100%) with steps (1)
- Correct atom economy for Pathway 2 (64.0%) with steps (1)

(ii)
- Mentions CO2 utilization/mitigating greenhouse effect/greenhouse gas reduction (1)
- Mentions toxic nature of CO / non-renewability of fossil sources for Pathway 1 OR potential carbon-neutrality with green hydrogen (1)

(iii) (1)
- Mentions alternative pathway with lower activation energy (1)

(iii) (2)
- Explains irreversible strong adsorption / blocking of active sites (1)
- Suggests removal of sulfur prior to reaction / using ZnO guard bed (1)

**Part (a)**
(i) Add \(\text{NaHCO}_3(aq)\) or \(\text{Na}_2\text{CO}_3(aq)\) to the sample. A carboxylic acid will react to produce carbon dioxide gas (bubbles), whereas an ester will not show any observable reaction.
(ii) (1) \(\text{C=O}\) (carbonyl) group.
(2) Carboxylic acids contain an \(\text{O-H}\) group which gives a broad absorption band in the region \(2500 - 3300\text{ cm}^{-1}\). The absence of this band rules out all carboxylic acid isomers (e.g., butanoic acid).
(iii) (1) \(\text{CH}_3\text{CO}^+\)
(2) Since \(A\) has the molecular formula \(C_4H_8O_2\) and is an ester, it must be methyl propanoate, ethyl ethanoate, propyl methanoate, or isopropyl methanoate. The base peak at \(m/z = 43\) corresponds to the acylium ion \(\text{CH}_3\text{CO}^+\), indicating the presence of a \(\text{CH}_3\text{CO}-\) group. The peak at \(m/z = 61\) is due to the protonated acid species \([\text{CH}_3\text{COOH}_2]^+\), which is formed by the elimination of ethene (mass 28) from the molecular ion of ethyl ethanoate (\(M^+ = 88\)). Therefore, the structural formula of \(A\) is \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) (ethyl ethanoate).

**Part (b)**
(i) \(\text{Cr}_2\text{O}_7^{2-}(aq)\) is orange-colored and absorbs visible light. Its absorbance is directly proportional to its concentration (obeys the Beer-Lambert Law).
(ii) A colorimeter measures the intensity of light of a specific wavelength passing through a sample solution compared to a reference blank. A complementary filter is chosen because it allows the wavelength of light that is most strongly absorbed by the analyte (blue light for orange dichromate) to pass through, maximizing the sensitivity of the measurement.
(iii) (1) \(C = \frac{A}{1540} = \frac{0.462}{1540} = 3.0 \times 10^{-4}\text{ mol dm}^{-3}\)
(2) Dilution factor = \(\frac{250.0\text{ cm}^3}{10.0\text{ cm}^3} = 25\)
Original concentration = \(3.0 \times 10^{-4} \times 25 = 7.5 \times 10^{-3}\text{ mol dm}^{-3}\)
(3) Colorimetry is much faster, has higher sensitivity (can measure lower concentrations), and is non-destructive.

**Part (c)**
(i) Draw a starting line using a pencil on the TLC plate and spot a small amount of the sample on this line. Place the plate vertically inside a developing chamber containing a suitable solvent, ensuring the solvent level is below the start line. Cover the chamber with a lid and allow the solvent to ascend near the top of the plate, then remove and mark the solvent front.
(ii) View the TLC plate under ultraviolet (UV) light, or place the plate in a sealed container with iodine crystals to visualize the spots.
(iii) \(R_f = \frac{3.6\text{ cm}}{8.0\text{ cm}} = 0.45\)

**Part (a) (8 marks)**
(i) Add \(\text{NaHCO}_3(aq)\) or \(\text{Na}_2\text{CO}_3(aq)\) to the sample. [1M]
Carboxylic acid gives bubbles (or effervescence), while ester gives no observable change. [1M]
(ii) (1) \(\text{C=O}\) stretching / carbonyl group / ester group [1M]
(2) Carboxylic acids contain an \(\text{O-H}\) group which absorbs in the region \(2500 - 3300\text{ cm}^{-1}\). [1M]
(iii) (1) \(\text{CH}_3\text{CO}^+\) (formula and charge must be correct) [1M]
(2)
- States that \(A\) is an ester and identifies possible isomers (methyl propanoate, ethyl ethanoate, etc.). [1M]
- Attributes \(m/z = 43\) to \(\text{CH}_3\text{CO}^+\) or explains the loss of 28 (ethene) from \(M^+\) (88) to form \(m/z = 61\) (protonated ethanoic acid), which indicates an ethyl ester. [1M]
- Correctly identifies \(A\) as ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\). [1M]

**Part (b) (7 marks)**
(i) \(\text{Cr}_2\text{O}_7^{2-}(aq)\) is colored / absorbs visible light, and its absorbance is proportional to its concentration. [1M]
(ii)
- A colorimeter measures the intensity of light passing through the sample and compares it to a blank. [1M]
- A complementary filter selects light of the wavelength most strongly absorbed by the analyte, maximizing sensitivity. [1M]
(iii) (1) \(C = 3.0 \times 10^{-4}\text{ mol dm}^{-3}\) [1M]
(2)
- Correct dilution factor of 25. [1M]
- Correct calculation of original concentration = \(7.5 \times 10^{-3}\text{ mol dm}^{-3}\) (with unit). [1M]
(3) Any one of the following: faster / non-destructive / higher sensitivity (can measure lower concentrations) / suitable for continuous monitoring. [1M]

**Part (c) (5 marks)**
(i)
- Draw a starting line with pencil and spot the sample on it. [1M]
- Place the plate in a development chamber, ensuring the solvent level is below the starting line. [1M]
- Cover with a lid, allow solvent to ascend and mark the solvent front. [1M]
(ii) Visualize under UV light / in an iodine chamber. [1M]
(iii) \(R_f = 0.45\) (accept \(0.45\) or \(9/20\), no unit) [1M]