2023 HKDSE Chemistry Practice Paper with Answers

Methanal (\(\text{H}_2\text{C=O}\)) has a trigonal planar geometry around the central carbon atom (bond angles \(\approx 120^\circ\)). Because the \(\text{C=O}\) bond dipole is significantly more polar than the \(\text{C}-\text{H}\) bond dipoles, the individual bond dipoles do not cancel out, resulting in a net dipole moment. \(\text{BF}_3\) is planar but completely symmetrical (trigonal planar), so its dipoles cancel (non-polar). \(\text{CH}_2\text{Cl}_2\) is polar but has a tetrahedral (non-planar) geometry. \(\text{C}_2\text{H}_4\) is planar but non-polar due to symmetry.

1 mark for option C.

Zinc is more reactive (higher in electrochemical series) than silver, so zinc acts as the anode (negative electrode) where oxidation occurs: \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\), decreasing the mass of the zinc electrode. Silver acts as the cathode (positive electrode) where reduction occurs: \(\text{Ag}^+\text{(aq)} + \text{e}^- \rightarrow \text{Ag(s)}\), increasing the mass of the silver electrode. Electrons flow externally from Zn to Ag. Nitrate anions in the salt bridge migrate towards the zinc half-cell to balance the accumulation of positive \(\text{Zn}^{2+}\) ions.

1 mark for option C.

Number of moles of \(\text{H}_2\text{Y} = 0.02500\text{ dm}^3 \times 0.0800\text{ mol dm}^{-3} = 0.00200\text{ mol}\).
The neutralisation reaction is: \(\text{H}_2\text{Y} + 2\text{NaOH} \rightarrow \text{Na}_2\text{Y} + 2\text{H}_2\text{O}\).
Moles of \(\text{NaOH}\) required \(= 2 \times 0.00200\text{ mol} = 0.00400\text{ mol}\).
Concentration of \(\text{NaOH} = \frac{0.00400\text{ mol}}{0.02000\text{ dm}^3} = 0.200\text{ mol dm}^{-3}\).

1 mark for option C.

(1) is correct: The carbon-carbon double bond (\(\text{C=C}\)) and primary alcohol group can both be oxidised by acidified \(\text{KMnO}_4\), decolourising the purple solution.
(2) is correct: Each carbon atom of the \(\text{C=C}\) double bond is attached to two different groups (C2 is bonded to -H and -\(\text{CH}_3\); C3 is bonded to -H and -\(\text{CH}_2\text{OH}\)), so it exhibits cis-trans isomerism.
(3) is incorrect: The hydroxyl group is on a terminal carbon (primary alcohol), so oxidation yields an aldehyde (but-2-enal) or a carboxylic acid (but-2-enoic acid), not a ketone.

1 mark for option A.

Adding more \(\text{O}_2\text{(g)}\) increases the concentration of a reactant, which increases the frequency of effective collisions and thus increases the rate of the forward reaction. According to Le Chatelier's principle, increasing reactant concentration shifts the position of equilibrium to the right, thereby increasing the equilibrium yield of \(\text{SO}_3\text{(g)}\).
- Increasing temperature increases rate but decreases yield for this exothermic reaction.
- Adding a catalyst increases rate but has no effect on equilibrium yield.
- Increasing container volume decreases concentrations, lowering reaction rates.

1 mark for option C.

Intermolecular hydrogen bonding occurs between molecules containing a hydrogen atom bonded directly to a highly electronegative atom (N, O, or F), which interacts with a lone pair on an electronegative atom of a neighbouring molecule.

- (1) Propan-2-ol contains an \(-\text{OH}\) group, so it can form intermolecular hydrogen bonds.
- (2) Ethanamide contains an \(-\text{NH}_2\) group and a \(\text{C=O}\) group, so it forms extensive intermolecular hydrogen bonds.
- (3) Methoxyethane is an ether with a \(\text{C-O-C}\) linkage. It has no hydrogen atom directly attached to an oxygen atom, so it cannot form intermolecular hydrogen bonds with other methoxyethane molecules.

Therefore, only (1) and (2) can form intermolecular hydrogen bonds among their own molecules.

A (1 mark): Identifies that both propan-2-ol and ethanamide possess \(-\text{OH}\) and \(-\text{NH}_2\) groups respectively, enabling intermolecular hydrogen bonding, whereas methoxyethane does not.

At the cathode, \(\text{H}^+(\text{aq})\) ions (from the auto-ionization of water) have a higher reduction potential than \(\text{Na}^+(\text{aq})\) ions and are preferentially discharged to form \(\text{H}_2(\text{g})\):

\[2\text{H}^+(\text{aq}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g})\]

or

\[2\text{H}_2\text{O}(\text{l}) + 2\text{e}^- \rightarrow \text{H}_2(\text{g}) + 2\text{OH}^-(\text{aq})\]

As \(\text{H}^+\) ions are consumed, the equilibrium \(\text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{OH}^-(\text{aq})\) shifts to the right, causing \([\text{OH}^-]\) to exceed \([\text{H}^+]\). Thus, the solution around the cathode becomes alkaline. Both statements are true, and the 2nd statement correctly explains the 1st statement.

A (1 mark): Both statements are true and the 2nd statement is the correct chemical reason for the alkalinity observed at the cathode.

Adding an unreactive noble gas (such as neon) to an equilibrium mixture at constant volume does not alter the volume of the container, nor does it affect the number of moles or partial pressures of \(\text{SO}_2\), \(\text{O}_2\), or \(\text{SO}_3\).

Since the concentrations of all reacting species remain unchanged, the reaction quotient \(Q_c\) remains equal to \(K_c\). Furthermore, \(K_c\) depends only on temperature, which is held constant. Therefore, the position of equilibrium remains unchanged.

D (1 mark): Recognizes that adding an inert gas at constant volume does not change partial concentrations of reactants/products, so equilibrium position is unaffected.

Step 1: Write the balanced chemical equation:
\[\text{H}_2\text{X}(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{X}(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\]

Step 2: Calculate the amount of \(\text{H}_2\text{X}\) reacted:
\[n(\text{H}_2\text{X}) = 0.120\,\text{mol dm}^{-3} \times \frac{25.0}{1000}\,\text{dm}^3 = 3.00 \times 10^{-3}\,\text{mol}\]

Step 3: Determine the amount of \(\text{NaOH}\) required based on mole ratio (\(1 : 2\)):
\[n(\text{NaOH}) = 2 \times 3.00 \times 10^{-3}\,\text{mol} = 6.00 \times 10^{-3}\,\text{mol}\]

Step 4: Calculate the molarity of \(\text{NaOH}\):
\[[\text{NaOH}] = \frac{6.00 \times 10^{-3}\,\text{mol}}{\frac{30.0}{1000}\,\text{dm}^3} = 0.200\,\text{mol dm}^{-3}\]

C (1 mark): Correctly applies the 1:2 stoichiometry for a diprotic acid with NaOH to calculate the concentration as \(0.200\,\text{mol dm}^{-3}\).

Acidified potassium dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7 / \text{H}^+\)) is an oxidizing agent that turns from orange (\(\text{Cr}_2\text{O}_7^{2-}\)) to green (\(\text{Cr}^{3+}\)) when reduced.

- (1) Butan-1-ol is a primary alcohol and is oxidized (orange to green). 2-Methylpropan-2-ol is a tertiary alcohol and is resistant to oxidation (remains orange). They can be distinguished.
- (2) Propanal is an aldehyde and is oxidized to propanoic acid (orange to green). Propanone is a ketone and is not readily oxidized (remains orange). They can be distinguished.
- (3) Ethanoic acid is a carboxylic acid and ethyl ethanoate is an ester; neither is oxidized by acidified dichromate under these conditions (both remain orange, showing no colour change). They cannot be distinguished.

Hence, only pairs (1) and (2) can be distinguished.

A (1 mark): Identifies that acidified potassium dichromate undergoes an orange-to-green colour change with primary alcohols and aldehydes, but does not react with tertiary alcohols, ketones, carboxylic acids, or esters.

Propanoic acid molecules can form stable hydrogen-bonded dimers via two hydrogen bonds per dimer, resulting in extensive intermolecular hydrogen bonding and the highest boiling point. Butan-1-ol also forms intermolecular hydrogen bonds via its hydroxyl group, but does not form dimer structures like carboxylic acids, so its boiling point is lower than that of propanoic acid. Methyl ethanoate is an ester and cannot form intermolecular hydrogen bonds with itself (it only has permanent dipole-dipole attractions and van der Waals' forces), so it has the lowest boiling point. Therefore, the order is (1) > (2) > (3).

A (1 mark): Identifies propanoic acid has the strongest intermolecular attractions due to dimer formation, followed by butan-1-ol (hydrogen bonding), and methyl ethanoate (dipole-dipole forces only).

Since nickel is higher in the electrochemical series than silver, nickel acts as the negative electrode (anode) and undergoes oxidation: \( \text{Ni}(\text{s}) \rightarrow \text{Ni}^{2+}(\text{aq}) + 2\text{e}^- \). Therefore, the nickel rod dissolves and its mass decreases (statement (2) is correct). Electrons flow from the anode (nickel rod) to the cathode (silver rod) through the external circuit (statement (1) is correct). In half-cell Y, \( \text{Ag}^+ \) ions are reduced to \( \text{Ag}(\text{s}) \), creating a deficit of positive charge; cations from the salt bridge (\( \text{K}^+ \)) migrate toward half-cell Y (cathode), not half-cell X (statement (3) is incorrect).

A (1 mark): Identifies statements (1) and (2) as correct while recognizing that cations migrate to the cathode (half-cell Y).

Molar mass of \( \text{BaSO}_4 = 137.3 + 32.1 + 4(16.0) = 233.4\text{ g mol}^{-1} \).
Number of moles of \( \text{BaSO}_4 = \frac{0.2334\text{ g}}{233.4\text{ g mol}^{-1}} = 1.00 \times 10^{-3}\text{ mol} \).
Thus, in \( 25.0\text{ cm}^3 \) of solution, moles of \( \text{H}_2\text{SO}_4 = 1.00 \times 10^{-3}\text{ mol} \).
Since each \( \text{H}_2\text{SO}_4 \) provides \( 2\text{ H}^+ \), moles of \( \text{H}^+ \) from \( \text{H}_2\text{SO}_4 = 2.00 \times 10^{-3}\text{ mol} \).
Total moles of \( \text{OH}^- \) required for neutralization \( = 0.0300\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 6.00 \times 10^{-3}\text{ mol} \).
Total moles of \( \text{H}^+ = 6.00 \times 10^{-3}\text{ mol} \).
Moles of \( \text{H}^+ \) from \( \text{HCl} = 6.00 \times 10^{-3} - 2.00 \times 10^{-3} = 4.00 \times 10^{-3}\text{ mol} \).
Concentration of \( \text{HCl} = \frac{4.00 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.160\text{ M} \).

C (1 mark): Correctly calculates sulfate content to deduce \( \text{H}^+ \) from \( \text{H}_2\text{SO}_4 \), subtracts from total \( \text{H}^+ \), and determines \( [\text{HCl}] = 0.160\text{ M} \).

(1) is correct: The compound contains a carbon-carbon double bond (and a secondary alcohol group), so it reacts with and decolourizes acidified \( \text{KMnO}_4(\text{aq}) \).
(2) is correct: Each carbon atom of the \( \text{C}=\text{C} \) double bond is attached to two different groups (C3 is attached to \( -\text{H} \) and \( -\text{CH(OH)CH}_3 \); C4 is attached to \( -\text{H} \) and \( -\text{CH}_3 \)), so it exhibits cis-trans isomerism.
(3) is correct: Carbon-2 is bonded to four different groups/atoms (\( -\text{H} \), \( -\text{OH} \), \( -\text{CH}_3 \), and \( -\text{CH}=\text{CHCH}_3 \)), which makes it a chiral carbon centre.

D (1 mark): Identifies all three statements (1), (2), and (3) as correct.

The forward reaction is exothermic (\( \Delta H < 0 \)). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium position in the forward (exothermic) direction, thereby increasing the equilibrium yield of \( \text{SO}_3(\text{g}) \). Furthermore, the equilibrium constant \( K_c \) is temperature-dependent only; for an exothermic reaction, a decrease in temperature increases \( K_c \).
- Changes in pressure/volume or reactant concentrations (A and B) alter equilibrium positions but do not change the value of \( K_c \).
- A catalyst (D) increases the rates of forward and backward reactions equally, without changing the equilibrium yield or \( K_c \).

C (1 mark): Understands the effect of temperature on exothermic equilibria and that \( K_c \) is affected only by temperature changes.

The compound has one chiral carbon centre (at C-2, bearing \(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH=CHCH}_3\)) and one carbon-carbon double bond capable of showing \(\text{cis}\)-\(\text{trans}\) (or \(E\)/\(Z\)) isomerism (at C-3, with different groups attached to each double-bonded carbon). Since the chiral centre gives 2 configurations (\(R\) and \(S\)) and the double bond gives 2 configurations (\(E\) and \(Z\)), the total number of stereoisomers is \(2 \times 2 = 4\): (2R, 3E), (2R, 3Z), (2S, 3E), and (2S, 3Z).

C (1 mark)

In \(\text{CH}_4\), the central carbon atom has 4 bond pairs and 0 lone pairs, giving a tetrahedral shape. In \(\text{NH}_4^+\), the central nitrogen atom also has 4 bond pairs and 0 lone pairs, giving a tetrahedral shape. Therefore, both species have the same shape (tetrahedral). For the other pairs: \(\text{CO}_2\) is linear while \(\text{SO}_2\) is bent (V-shaped); \(\text{NH}_3\) is trigonal pyramidal while \(\text{BF}_3\) is trigonal planar; \(\text{H}_2\text{O}\) is bent (V-shaped) while \(\text{BeCl}_2\) is linear.

C (1 mark)

The value of the equilibrium constant \(K_c\) for a given reaction depends solely on temperature. Because the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the position of equilibrium to the right (favouring the forward reaction according to Le Chatelier's principle), which increases the equilibrium concentration of \(\text{N}_2\text{O}_4\text{(g)}\) relative to \(\text{NO}_2\text{(g)}\), thereby increasing \(K_c\). Changes in concentration, volume, or catalysts do not alter \(K_c\).

B (1 mark)

Zinc is more reactive than copper, so zinc acts as the anode (negative electrode) where oxidation occurs: \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\). Copper acts as the cathode (positive electrode) where reduction occurs: \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\). Electrons flow from Zn to Cu in the external wire. As \(\text{Zn}^{2+}\) ions accumulate in the anode compartment, nitrate ions (anions, \(\text{NO}_3^-\)) from the salt bridge migrate into the zinc half-cell to maintain electrical neutrality.

C (1 mark)

Reaction equation: \(\text{H}_2\text{X} + 2\text{KOH} \rightarrow \text{K}_2\text{X} + 2\text{H}_2\text{O}\). Number of moles of \(\text{H}_2\text{X} = 0.120\text{ mol dm}^{-3} \times \frac{25.00}{1000}\text{ dm}^3 = 3.00 \times 10^{-3}\text{ mol}\). According to the mole ratio (1 : 2), number of moles of \(\text{KOH} = 2 \times (3.00 \times 10^{-3}\text{ mol}) = 6.00 \times 10^{-3}\text{ mol}\). Molarity of \(\text{KOH} = \frac{6.00 \times 10^{-3}\text{ mol}}{\frac{18.75}{1000}\text{ dm}^3} = 0.320\text{ M}\).

C (1 mark)

1. \( \text{CF}_4 \) has a regular tetrahedral shape with four identical polar \( \text{C}-\text{F} \) bonds. The individual bond dipoles cancel out symmetrically, resulting in a dipole moment of zero (non-polar).
2. \( \text{NF}_3 \) has a trigonal pyramidal shape due to one lone pair on the nitrogen atom. The bond dipoles do not cancel out, so it has a net dipole moment (polar).
3. \( \text{OF}_2 \) has a V-shaped (bent) geometry due to two lone pairs on the oxygen atom. The bond dipoles do not cancel out, so it has a net dipole moment (polar).

Therefore, only (1) has a dipole moment of zero.

Correct option [A] awards 1 mark. Incorrect options award 0 marks.

- 1st statement is TRUE: When \( \text{NaOH(aq)} \) is added dropwise to \( \text{Al(NO}_3)_3\text{(aq)} \), a white precipitate of aluminium hydroxide forms: \( \text{Al}^{3+}\text{(aq)} + 3\text{OH}^-\text{(aq)} \rightarrow \text{Al(OH)}_3\text{(s)} \). Upon adding excess \( \text{NaOH(aq)} \), the precipitate dissolves to give a colourless solution containing tetrahydroxidoaluminate ions: \( \text{Al(OH)}_3\text{(s)} + \text{OH}^-\text{(aq)} \rightarrow [\text{Al(OH)}_4]^-\text{(aq)} \).
- 2nd statement is TRUE: Aluminium hydroxide exhibits amphoteric behaviour, reacting with both strong acids and strong alkalis.
- Since the dissolution in excess alkali is a direct consequence of the amphoteric nature of aluminium hydroxide, the 2nd statement is a correct explanation of the 1st statement.

Correct option [A] awards 1 mark. Incorrect options award 0 marks.

1. Write the balanced chemical equation:
\[ \text{H}_2\text{X(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{X(aq)} + 2\text{H}_2\text{O(l)} \]

2. Calculate the amount of \( \text{NaOH} \) used:
\[ n(\text{NaOH}) = 0.200\,\text{mol dm}^{-3} \times \frac{30.0}{1000}\,\text{dm}^3 = 6.00 \times 10^{-3}\,\text{mol} \]

3. Calculate the amount of \( \text{H}_2\text{X} \) in \( 25.0\,\text{cm}^3 \):
\[ n(\text{H}_2\text{X})_{\text{in } 25.0\,\text{cm}^3} = \frac{6.00 \times 10^{-3}\,\text{mol}}{2} = 3.00 \times 10^{-3}\,\text{mol} \]

4. Calculate the amount of \( \text{H}_2\text{X} \) in \( 250.0\,\text{cm}^3 \):
\[ n(\text{H}_2\text{X})_{\text{in } 250.0\,\text{cm}^3} = 3.00 \times 10^{-3}\,\text{mol} \times \frac{250.0}{25.0} = 3.00 \times 10^{-2}\,\text{mol} \]

5. Calculate the mass of \( \text{H}_2\text{X} \):
\[ m = 3.00 \times 10^{-2}\,\text{mol} \times 126.0\,\text{g mol}^{-1} = 3.78\,\text{g} \]

Correct option [C] awards 1 mark. Incorrect options award 0 marks.

- (1) Decreasing the volume increases the total pressure. According to Le Chatelier's Principle, the system responds by shifting towards the side with fewer moles of gas (from 3 moles on the left to 2 moles on the right). Thus, equilibrium shifts to the right. (Correct)
- (2) Adding an inert gas (He) at constant volume increases the total pressure, but does not alter the concentrations or partial pressures of the reacting gases. Therefore, there is no shift in the equilibrium position. (Incorrect)
- (3) The forward reaction is exothermic (\( \Delta H < 0 \)). Lowering the temperature causes the equilibrium to shift in the heat-producing (forward/right) direction to oppose the change. (Correct)

Hence, (1) and (3) only are correct.

Correct option [B] awards 1 mark. Incorrect options award 0 marks.

- **A (\( \text{CH}_3\text{CH}=\text{CHCH(OH)CH}_3 \)):**
- Around the double bond (\( \text{C}=\text{C} \)), the left carbon is attached to \( -\text{H} \) and \( -\text{CH}_3 \), and the right carbon is attached to \( -\text{H} \) and \( -\text{CH(OH)CH}_3 \). Since each carbon atom of the double bond bears two different groups, it exhibits cis-trans isomerism.
- The carbon atom bonded to the \( -\text{OH} \) group is bonded to four different groups: \( -\text{H} \), \( -\text{OH} \), \( -\text{CH}_3 \), and \( -\text{CH}=\text{CHCH}_3 \). It is a chiral carbon, so it exhibits enantiomerism.
- Thus, compound A exhibits both types of isomerism.
- **B (\( \text{CH}_2=\text{CHCH(OH)CH}_3 \)):** Contains a chiral carbon, but the terminal double bond carbon has two identical hydrogen atoms, so it cannot show cis-trans isomerism.
- **C (\( \text{CH}_3\text{CH}=\text{C(CH}_3)_2 \)):** One of the double-bonded carbons has two identical methyl groups, so no cis-trans isomerism; also lacks a chiral carbon.
- **D (\( \text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3 \)):** Exhibits cis-trans isomerism, but has no chiral carbon center.

Correct option [A] awards 1 mark. Incorrect options award 0 marks.

Statement (1) is correct: \(\text{BeCl}_2\) has a linear shape with a bond angle of \(180^\circ\), whereas \(\text{SO}_2\) has a bent shape with a bond angle of approximately \(119^\circ\).
Statement (2) is correct: The central S atom in \(\text{SO}_2\) has 2 bonding domains and 1 lone pair (bent shape); the central O atom in \(\text{OF}_2\) has 2 bonding pairs and 2 lone pairs (bent shape).
Statement (3) is incorrect: \(\text{SO}_2\) is a bent molecule with polar \(\text{S}=\text{O}\) bonds; the dipole moments do not cancel out, so \(\text{SO}_2\) is a polar molecule.

A (1 mark): Statements (1) and (2) only are correct.

Since \(E^\circ(\text{Ag}^+/\text{Ag}) = +0.80\text{ V} > E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}\):
- The silver half-cell acts as the cathode (reduction): \(\text{Ag}^+(\text{aq}) + \text{e}^- \rightarrow \text{Ag}(\text{s})\).
- The platinum half-cell acts as the anode (oxidation): \(\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{e}^-\).

Evaluating options:
- A is incorrect: Electrons flow from the anode (platinum electrode) to the cathode (silver electrode) in the external circuit.
- B is correct: \(\text{Fe}^{2+}\) is oxidised to \(\text{Fe}^{3+}\), so \([\text{Fe}^{3+}(\text{aq})]\) increases.
- C is incorrect: The platinum electrode is inert and its mass remains unchanged.
- D is incorrect: Adding \(\text{NaCl}(\text{aq})\) precipitates \(\text{Ag}^+\) as \(\text{AgCl}(\text{s})\), decreasing \([\text{Ag}^+(\text{aq})]\), which lowers the reduction potential of the cathode and decreases the cell voltage.

B (1 mark): The concentration of \(\text{Fe}^{3+}(\text{aq})\) in half-cell 1 increases.

Equation: \(\text{H}_2\text{A}(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{A}(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\)

1. Moles of \(\text{NaOH}\) used in titration:
\(n(\text{NaOH}) = 0.100\text{ mol dm}^{-3} \times \frac{20.00}{1000}\text{ dm}^3 = 2.00 \times 10^{-3}\text{ mol}\)

2. Moles of \(\text{H}_2\text{A}\) in \(25.00\text{ cm}^3\):
\(n(\text{H}_2\text{A}) = \frac{1}{2} \times n(\text{NaOH}) = \frac{1}{2} \times 2.00 \times 10^{-3} = 1.00 \times 10^{-3}\text{ mol}\)

3. Moles of \(\text{H}_2\text{A}\) in \(250.0\text{ cm}^3\):
\(n_{\text{total}}(\text{H}_2\text{A}) = 1.00 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.00} = 1.00 \times 10^{-2}\text{ mol}\)

4. Mass of pure \(\text{H}_2\text{A}\):
\(m = 1.00 \times 10^{-2}\text{ mol} \times 126.0\text{ g mol}^{-1} = 1.26\text{ g}\)

5. Percentage by mass:
\(\text{Percentage} = \frac{1.26\text{ g}}{1.50\text{ g}} \times 100\% = 84.0\%\)

C (1 mark): 84.0%

Statement (1) is correct: The \(\text{C}=\text{C}\) double bond has two different groups attached to each double-bonded carbon (\(-\text{H}\) and \(-\text{CH}_3\) on one carbon; \(-\text{H}\) and \(-\text{CH}(\text{OH})\text{CH}_3\) on the other), so it exhibits cis-trans isomerism.
Statement (2) is correct: Carbon-4 is bonded to four different groups (\(-\text{H}\), \(-\text{OH}\), \(-\text{CH}_3\), and \(-\text{CH}=\text{CHCH}_3\)), making it a chiral carbon center. Hence, it exhibits enantiomerism.
Statement (3) is correct: The hydroxyl group is on a secondary carbon atom (secondary alcohol), which can be oxidised to a ketone by acidified potassium dichromate solution, reducing \(\text{Cr}_2\text{O}_7^{2-}\) (orange) to \(\text{Cr}^{3+}\) (green).

D (1 mark): Statements (1), (2) and (3) are all correct.

The value of the equilibrium constant \(K_c\) depends ONLY on temperature.
Since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature shifts the equilibrium position to the right (forward direction) according to Le Chatelier's principle. This increases the concentration of \(\text{N}_2\text{O}_4(\text{g})\) and decreases the concentration of \(\text{NO}_2(\text{g})\) at the new equilibrium, resulting in an increase in \(K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}\).

Changing pressure/volume, adding inert gas at constant volume, or adding a catalyst does not alter the numerical value of \(K_c\).

C (1 mark): Decreasing the temperature of the reaction system.

To determine the shape of each molecule:
- (1) \( \text{SO}_3 \): The central sulphur atom forms three electron domains (three bonding domains, zero lone pairs). The electron pair geometry and molecular geometry are both trigonal planar (planar triangular) with bond angles of \(120^\circ\).
- (2) \( \text{NF}_3 \): The central nitrogen atom has four electron domains (three bonding pairs and one lone pair). The molecular geometry is trigonal pyramidal.
- (3) \( \text{BF}_3 \): The central boron atom forms three bonding pairs with no lone pairs. The molecular geometry is planar triangular with bond angles of \(120^\circ\).

Therefore, (1) and (3) only have a planar triangular shape.

1 mark for option C. Options A, B, and D are incorrect.

Since metal \( \text{M} \) is higher in the electrochemical series (more reactive) than silver:
- Electrode \( \text{M} \) acts as the anode (oxidation): \( \text{M(s)} \rightarrow \text{M}^{2+}\text{(aq)} + 2\text{e}^- \).
For \( 0.020\text{ mol} \) of electrons transferred, the moles of \( \text{M}^{2+} \) formed = \( \frac{0.020}{2} = 0.010\text{ mol} \). Hence, statement B is correct.
- Silver electrode acts as the cathode (reduction): \( \text{Ag}^+\text{(aq)} + \text{e}^- \rightarrow \text{Ag(s)} \).
The mass of the silver electrode increases (not decreases) by \( 0.020\text{ mol} \times 107.9\text{ g mol}^{-1} = 2.16\text{ g} \). Statement A is incorrect.
- \( \text{K}^+ \) cations from the salt bridge migrate towards the cathode (silver half-cell) to balance the loss of \( \text{Ag}^+ \) ions, not towards electrode \( \text{M} \). Statement C is incorrect.
- Reduction (not oxidation) takes place at the silver electrode. Statement D is incorrect.

1 mark for option B. Options A, C, and D are incorrect.

Step 1: Calculate moles of \( \text{H}^+ \) provided by \( \text{H}_2\text{SO}_4 \):
\( n(\text{H}^+) = 2 \times c(\text{H}_2\text{SO}_4) \times V = 2 \times 0.120\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 0.00600\text{ mol} \)

Step 2: Calculate moles of \( \text{OH}^- \) provided by \( \text{NaOH} \):
\( n(\text{OH}^-) = c(\text{NaOH}) \times V = 0.200\text{ mol dm}^{-3} \times 0.0350\text{ dm}^3 = 0.00700\text{ mol} \)

Step 3: Determine excess moles of \( \text{OH}^- \):
\( n(\text{OH}^-)_{\text{excess}} = 0.00700 - 0.00600 = 0.00100\text{ mol} \)

Step 4: Calculate total volume and concentration:
\( V_{\text{total}} = 25.0 + 35.0 = 60.0\text{ cm}^3 = 0.0600\text{ dm}^3 \)
\( [\text{OH}^-] = \frac{0.00100\text{ mol}}{0.0600\text{ dm}^3} = 0.0167\text{ mol dm}^{-3} \)

1 mark for option A. Options B, C, and D are common calculation errors (e.g. failing to account for the dibasic nature of sulfuric acid).

Deduction from the test observations:
1. Turning acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \) from orange to green indicates that \( \text{X} \) can be oxidized (contains a primary/secondary alcohol or aldehyde group).
2. No reaction with \( \text{NaHCO}_3\text{(aq)} \) means \( \text{X} \) is not a carboxylic acid. This eliminates \( \text{CH}_3\text{CH}_2\text{CH}_2\text{COOH} \) (Option A).
3. Reaction with \( \text{Na(s)} \) to liberate \( \text{H}_2\text{(g)} \) indicates the presence of a hydroxyl group (\( -\text{OH} \)). This eliminates esters such as \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \) (Option B) and \( \text{HCOOCH}_2\text{CH}_2\text{CH}_3 \) (Option D).

Structure C, \( \text{CH}_3\text{COCH(OH)CH}_3 \) (3-hydroxybutan-2-one), has formula \( \text{C}_4\text{H}_8\text{O}_2 \), contains a secondary alcohol group which reacts with \( \text{Na} \) and is oxidized by acidified dichromate, and lacks a carboxyl group, fitting all observations.

1 mark for option C. Options A, B, and D are incorrect based on functional group chemical tests.

- 1st statement: \( \text{N}_2\text{O}_4\text{(g)} \) is colourless and \( \text{NO}_2\text{(g)} \) is brown. The forward reaction is endothermic (\( \Delta H > 0 \)). When temperature increases, the system shifts in the endothermic direction (to the right) according to Le Chatelier's principle, increasing the concentration of \( \text{NO}_2\text{(g)} \) and making the gas darker brown. (True)
- 2nd statement: For an endothermic reaction, raising temperature increases the value of the equilibrium constant \( K_c \). (True)
- Explanation link: Because \( K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \) increases as temperature rises, the equilibrium position lies further to the right, yielding a higher equilibrium concentration of \( \text{NO}_2\text{(g)} \) and thus a darker colour. Hence, the 2nd statement is the correct explanation of the 1st statement.

1 mark for option A. Options B, C, and D are incorrect.

The equation for the standard enthalpy change of formation of ethanoic acid is:
$$2\text{C(graphite)} + 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{COOH(l)}$$

Applying Hess's Law using standard enthalpy changes of combustion ($\Delta H^\circ_\text{c}$):
$$\begin{align*}
\Delta H^\circ_\text{f}[\text{CH}_3\text{COOH(l)}] &= 2 \times \Delta H^\circ_\text{c}[\text{C(graphite)}] + 2 \times \Delta H^\circ_\text{c}[\text{H}_2\text{(g)}] - \Delta H^\circ_\text{c}[\text{CH}_3\text{COOH(l)}] \\
&= 2(-393.5\text{ kJ mol}^{-1}) + 2(-285.8\text{ kJ mol}^{-1}) - (-874.2\text{ kJ mol}^{-1}) \\
&= -787.0 - 571.6 + 874.2 \\
&= -484.4\text{ kJ mol}^{-1}
\end{align*}$$

Thus, the correct answer is A.

A (1 mark): Correct calculation using Hess's Law: $\Delta H^\circ_\text{f} = 2(-393.5) + 2(-285.8) - (-874.2) = -484.4\text{ kJ mol}^{-1}$.

(a) Pent-2-ene has the molecular formula \(\text{C}_5\text{H}_{10}\). Upon catalytic hydrogenation with \(\text{H}_2/\text{Pt}\), it yields pentane. Since each carbon in the \(\text{C}=\text{C}\) double bond is attached to two different groups (\(-\text{H}\) and \(-\text{CH}_3\) on C2; \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\) on C3), it exhibits cis-trans isomerism.

(b) cis-pent-2-ene has the two hydrogen atoms on the same side of the double bond: \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2\text{CH}_3\). trans-pent-2-ene has the two hydrogen atoms on opposite sides of the double bond.

(c) The requirements for cis-trans isomerism are: (1) Restricted rotation about the carbon-carbon double bond (\(\text{C}=\text{C}\)), and (2) Each carbon atom involved in the double bond must be bonded to two different atoms or groups.

(a) Pent-2-ene (1 mark)
(b) Correct structural drawings showing stereochemistry:
- cis-pent-2-ene: with \(-\text{H}\) atoms on the same side of \(\text{C}=\text{C}\) (1 mark)
- trans-pent-2-ene: with \(-\text{H}\) atoms on opposite sides of \(\text{C}=\text{C}\) (1 mark)
(c) Restricted rotation about the \(\text{C}=\text{C}\) bond AND each carbon atom of the double bond is attached to two different groups / atoms. (1 mark)

(a) (i) \(\text{BF}_3\) has 3 bond pairs and 0 lone pairs on boron. Its 3D drawing shows a flat trigonal planar arrangement with bond angles of \(120^\circ\).
(ii) \(\text{NF}_3\) has 3 bond pairs and 1 lone pair on nitrogen. Its 3D drawing uses wedges and dashes with one lone pair shown on N.

(b) The shape of \(\text{NF}_3\) is trigonal pyramidal.

(c) Although both \(\text{B}-\text{F}\) and \(\text{N}-\text{F}\) bonds are polar due to electronegativity differences, \(\text{BF}_3\) is symmetrical (trigonal planar), causing the individual bond dipoles to cancel each other out (dipole moment = 0). In contrast, \(\text{NF}_3\) has an asymmetrical (trigonal pyramidal) geometry, so the individual bond dipoles do not cancel out, giving a net dipole moment.

(a)
(i) Correct 3D representation of \(\text{BF}_3\) (trigonal planar, \(120^\circ\)) (1 mark)
(ii) Correct 3D representation of \(\text{NF}_3\) (showing wedges/dashes and 1 lone pair on N) (1 mark)
(b) Trigonal pyramidal (1 mark)
(c) Symmetrical shape of \(\text{BF}_3\) leads to cancellation of bond dipoles, whereas the asymmetrical shape / presence of lone pair in \(\text{NF}_3\) prevents cancellation of bond dipoles, resulting in a net dipole moment. (1 mark)

(a) The \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell has a more positive standard reduction potential (\(+0.77\text{ V}\)) than the \(\text{I}_2/\text{I}^-\)\ half-cell (\(+0.54\text{ V}\)). Therefore, \(\text{Fe}^{3+}\) undergoes reduction at the cathode, and \(\text{I}^-\) undergoes oxidation at the anode.
Reduction: \(2\text{Fe}^{3+}(\text{aq}) + 2e^- \rightarrow 2\text{Fe}^{2+}(\text{aq})\)
Oxidation: \(2\text{I}^-(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + 2e^-\)
Overall equation: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)

(b) Electrons flow from the negative electrode (anode, platinum electrode in \(\text{I}_2/\text{I}^-\)) to the positive electrode (cathode, platinum electrode in \(\text{Fe}^{3+}/\text{Fe}^{2+}\)) through the external wire.

(c) The salt bridge: (1) Completes the electrical circuit by allowing ions to migrate between the two half-cells, and (2) Maintains electrical neutrality in both solutions by supplying cations and anions to balance charges.

(a) \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\) (1 mark)
[State symbols not strictly required unless specified, but balancing must be correct]
(b) From the platinum electrode in the \(\text{I}_2/\text{I}^-\) half-cell to the platinum electrode in the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell. (1 mark)
(c)
- To complete the electrical circuit / allow the flow of ions. (1 mark)
- To maintain electrical neutrality in the two half-cells. (1 mark)

(a) Step 1 involves the dehydrohalogenation of 1-chloropropane to form propene (\(\mathbf{Y}\)). Reagent and condition: potassium hydroxide in ethanol (ethanolic \(\text{KOH}\)), heating under reflux.
(b) The intermediate \(\mathbf{Y}\) is propene: \(\text{CH}_3\text{CH}=\text{CH}_2\).
(c) Step 2 is the acid-catalysed hydration of propene to propan-2-ol (Markovnikov addition). Reagent and condition: \(\text{H}_2\text{O}\) with concentrated \(\text{H}_2\text{SO}_4\), heating / warming (or concentrated \(\text{H}_2\text{SO}_4\) followed by boiling with \(\text{H}_2\text{O}\)).
(d) Step 1 is an elimination reaction.

(a) \(\text{KOH}\) / \(\text{NaOH}\) in ethanol (ethanolic \(\text{KOH}\)), heat / heat under reflux (1 mark)
(b) \(\text{CH}_3\text{CH}=\text{CH}_2\) (1 mark)
(c) \(\text{H}_2\text{O}\) / dilute \(\text{H}_2\text{SO}_4\), heat (or concentrated \(\text{H}_2\text{SO}_4\) followed by \(\text{H}_2\text{O}\), heat) (1 mark)
(d) Elimination (1 mark)

(a) Number of moles of \(\text{NaOH}\) in \(20.00\text{ cm}^3\):
\(n(\text{NaOH}) = 0.125\text{ mol dm}^{-3} \times \frac{20.00}{1000}\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}\)

Equation for neutralisation:
\(\text{H}_2\text{A} + 2\text{NaOH} \rightarrow \text{Na}_2\text{A} + 2\text{H}_2\text{O}\)

Number of moles of \(\text{H}_2\text{A}\) in \(25.00\text{ cm}^3\):
\(n(\text{H}_2\text{A})_{25\text{ cm}^3} = \frac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3}\text{ mol}\)

Number of moles of \(\text{H}_2\text{A}\) in the whole \(100.0\text{ cm}^3\) solution:
\(n(\text{H}_2\text{A})_{\text{total}} = 1.25 \times 10^{-3}\text{ mol} \times \frac{100.0}{25.00} = 5.00 \times 10^{-3}\text{ mol}\)

Molar mass of \(\text{H}_2\text{A}\):
\(M = \frac{\text{mass}}{n} = \frac{0.590\text{ g}}{5.00 \times 10^{-3}\text{ mol}} = 118\text{ g mol}^{-1}\)

(b) Formula: \(\text{HOOC(CH}_2)_n\text{COOH}\)
Molar mass = \(2 \times (1.0 + 12.0 + 32.0) + n \times (12.0 + 2.0) = 2 \times 45.0 + 14.0n = 90.0 + 14.0n\)
\(90.0 + 14.0n = 118\)
\(14.0n = 28.0 \implies n = 2\)

(c) The titration is between a weak acid (dicarboxylic acid) and a strong base (\(\text{NaOH}\)), so the pH at the equivalence point is greater than 7 (alkaline). Methyl orange has a pH working range in the acidic region (pH 3.1–4.4) and would change colour prematurely before the equivalence point is reached.

(a)
- Calculation of total moles of \(\text{H}_2\text{A} = 5.00 \times 10^{-3}\text{ mol}\) (1 mark)
- Molar mass = \(118\text{ g mol}^{-1}\) (unit required) (1 mark)
(b) \(90.0 + 14.0n = 118 \implies n = 2\) (1 mark)
(c) The titration involves a weak acid and a strong base, so the equivalence point lies in the alkaline/basic pH region (\(\text{pH} > 7\)), whereas methyl orange has a pH transition range in the acidic region (\(\text{pH } 3.1 - 4.4\)). (1 mark)

(a) 2-Bromobutane has a chiral carbon (C2) bonded to four different groups: \(-\text{H}\), \(-\text{CH}_3\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{Br}\). The enantiomers are drawn as mirror images using solid, dashed, and wedged lines around the central carbon.

(b)(i) trans-But-2-ene has the two methyl groups (\(-\text{CH}_3\)) on opposite sides of the plane of the \(\text{C}=\text{C}\) double bond.

(ii) But-2-ene exhibits cis-trans isomerism because the \(\text{C}=\text{C}\) double bond has restricted (hindered) rotation due to the \(\pi\) bond, and each of the two carbon atoms forming the double bond is attached to two different groups (a hydrogen atom and a methyl group).

(a)
- Correct 3D wedge-dash representation for one enantiomer showing 4 different groups attached to the central carbon (1)
- Correct mirror-image structure drawn as the other enantiomer (1)
[Note: Deduct 1 mark if connectivity of ethyl group is shown incorrectly, e.g., bonded via \(\text{CH}_3\)]

(b)(i)
- Correct structural/displayed drawing of trans-but-2-ene showing \(-\text{CH}_3\) on opposite sides (1)

(b)(ii)
- Restricted rotation around the \(\text{C}=\text{C}\) bond AND each carbon atom of the double bond is bonded to two different atoms/groups (1)

(a) Phosphorus has 5 valence electrons and each fluorine has 7 valence electrons. Phosphorus shares one electron pair with each of the three fluorine atoms, leaving one lone pair on phosphorus. Each fluorine has 3 non-bonding pairs (lone pairs).

(b) The phosphorus atom has 4 electron pairs in its valence shell (3 bonding pairs and 1 lone pair). These electron pairs repel each other to achieve maximum separation / minimum repulsion, adopting a tetrahedral arrangement of electron pairs. Because one position is occupied by a lone pair (which exerts greater repulsion than bonding pairs), the molecular geometry / shape is trigonal pyramidal.

(c) Fluorine is more electronegative than phosphorus, so each \(\text{P}-\text{F}\) bond is polar with a bond dipole pointing towards fluorine. Since the molecule has an asymmetric / trigonal pyramidal shape, the bond dipoles do not cancel each other out, resulting in a net molecular dipole moment (polar molecule).

(a)
- Correct electron diagram with 3 single covalent \(\text{P}-\text{F}\) pairs, 1 lone pair on P, and 3 lone pairs on each F (1)

(b)
- Trigonal pyramidal (1)
- Mentions 4 electron pairs / 3 bond pairs + 1 lone pair around the central P atom repelling each other to minimize repulsion (1)

(c)
- \(\text{P}-\text{F}\) bonds are polar / electronegativity difference between P and F, and dipoles do not cancel out due to the non-symmetrical / pyramidal shape (1)

Step 1:
- Reagents & conditions: \(\text{H}_2\text{O}\) in the presence of concentrated \(\text{H}_2\text{SO}_4\) (or dilute \(\text{H}_2\text{SO}_4\)), heat.
- Intermediate organic product: Propan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_3\).

Step 2:
- Reagents & conditions: Acidified potassium dichromate solution (\(\text{K}_2\text{Cr}_2\text{O}_7\text{ / }\text{H}_2\text{SO}_4\text{(aq)}\)) or acidified potassium permanganate solution (\(\text{KMnO}_4\text{ / }\text{H}_2\text{SO}_4\text{(aq)}\)), heat under reflux.
- Final organic product: Propan-2-one, \(\text{CH}_3\text{COCH}_3\).

Step 1:
- Reagents & conditions: \(\text{H}_2\text{O}\) / \(\text{H}_2\text{SO}_4\text{(aq)}\), heat (or conc. \(\text{H}_2\text{SO}_4\) followed by \(\text{H}_2\text{O}\)) (1)
- Structural formula of intermediate: \(\text{CH}_3\text{CH(OH)CH}_3\) / propan-2-ol (1)

Step 2:
- Reagents & conditions: Acidified \(\text{K}_2\text{Cr}_2\text{O}_7\text{(aq)}\) / \(\text{H}^+\) (or acidified \(\text{KMnO}_4\)), heat (under reflux) (1)
- Structural formula of target product: \(\text{CH}_3\text{COCH}_3\) (1)
[Note: Maximum 4 marks. No marks awarded for a step if incorrect reagent produces an unworkable intermediate.]

(a) Molar mass of \(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O} = 2(1.0) + 2(12.0) + 4(16.0) + 2(18.0) = 126.0\text{ g mol}^{-1}\).
Number of moles of \(\text{H}_2\text{C}_2\text{O}_4 = \frac{1.26\text{ g}}{126.0\text{ g mol}^{-1}} = 0.0100\text{ mol}\).
Molarity of the solution \(= \frac{0.0100\text{ mol}}{0.2500\text{ dm}^3} = 0.0400\text{ mol dm}^{-3}\).

(b) Number of moles of \(\text{H}_2\text{C}_2\text{O}_4\) in \(25.00\text{ cm}^3 = 0.0400 \times \frac{25.00}{1000} = 1.00 \times 10^{-3}\text{ mol}\).
From the equation, mole ratio of \(\text{NaOH} : \text{H}_2\text{C}_2\text{O}_4 = 2 : 1\).
Number of moles of \(\text{NaOH} = 2 \times 1.00 \times 10^{-3} = 2.00 \times 10^{-3}\text{ mol}\).
Molarity of \(\text{NaOH} = \frac{2.00 \times 10^{-3}\text{ mol}}{0.02000\text{ dm}^3} = 0.100\text{ mol dm}^{-3}\).

(c) Phenolphthalein is suitable for titrating a weak acid (ethanedioic acid) with a strong base (sodium hydroxide). The end-point colour change is from colourless to pale pink.

(a)
- Molar mass calculation: \(126.0\text{ g mol}^{-1}\) and mole calculation: \(0.0100\text{ mol}\) (1)
- Molarity = \(0.0400\text{ mol dm}^{-3}\) (1)

(b)
- Molarity of \(\text{NaOH} = 0.100\text{ mol dm}^{-3}\) (with method / mole ratio 2:1 shown) (1)

(c)
- Phenolphthalein: colourless to (pale) pink (1)
[Reject: Methyl orange]

(a) Zinc has a higher position in the electrochemical series (or reactivity series) than copper, meaning zinc loses electrons more readily (is more easily oxidised). Thus, electrons flow through the external wire from the zinc electrode (anode) to the copper electrode (cathode).

(b)(i) At the anode (negative electrode), oxidation occurs:
\[\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\]
(ii) At the cathode (positive electrode), reduction occurs:
\[\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\]

(c) The salt bridge completes the electrical circuit and maintains electrical neutrality in the two half-cells by allowing migration of ions (cations to the cathode compartment and anions to the anode compartment).

(a)
- From Zn to Cu (through external circuit) AND Zn is more reactive / higher in electrochemical series than Cu (1)

(b)(i)
- \(\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^-\) (1)

(b)(ii)
- \(\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)}\) (1)
[State symbols not strictly required unless requested]

(c)
- Complete the electrical circuit / maintain electrical neutrality in the electrolyte solutions by ion migration (1)

(a) Chemical equation:
\(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)

(b) Indicator color change:
Since the acid in the conical flask is being titrated with \(\text{NaOH}\) from the burette, the initial solution is acidic (colourless) and turns pale pink at the permanent end point.

(c) Number of moles of \(\text{NaOH}\) in \(22.40\text{ cm}^3\):
\(n(\text{NaOH}) = 0.125\text{ mol dm}^{-3} \times \frac{22.40}{1000}\text{ dm}^3 = 2.80 \times 10^{-3}\text{ mol}\)

Reaction: \(\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}\)
Number of moles of unreacted \(\text{HCl}\) in \(25.0\text{ cm}^3\) aliquot = \(2.80 \times 10^{-3}\text{ mol}\)
Total number of moles of unreacted \(\text{HCl}\) in \(250.0\text{ cm}^3\) Solution \(\mathbf{S}\):
\(n(\text{HCl})_{\text{unreacted}} = 2.80 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.0} = 0.0280\text{ mol}\)

(d) Initial moles of \(\text{HCl}\) added:
\(n(\text{HCl})_{\text{initial}} = 2.00\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.100\text{ mol}\)

Moles of \(\text{HCl}\) reacted with \(\text{CaCO}_3\):
\(n(\text{HCl})_{\text{reacted}} = 0.100 - 0.0280 = 0.0720\text{ mol}\)

According to the stoichiometry \(\text{CaCO}_3 : \text{HCl} = 1 : 2\):
\(n(\text{CaCO}_3) = \frac{0.0720}{2} = 0.0360\text{ mol}\)

Molar mass of \(\text{CaCO}_3 = 40.1 + 12.0 + 3 \times 16.0 = 100.1\text{ g mol}^{-1}\)
Mass of \(\text{CaCO}_3 = 0.0360\text{ mol} \times 100.1\text{ g mol}^{-1} = 3.6036\text{ g}\)

Percentage by mass of \(\text{CaCO}_3\) in seashell sample:
\(\text{Percentage} = \frac{3.6036\text{ g}}{2.40\text{ g}} \times 100\% = 75.1\%\) (or \(75.15\%\))
(Wait, checking mass: \(3.6036 / 4.80\) vs sample size: mass of sample was \(4.80\text{ g}\) originally, but with \(2.40\text{ g}\) it would exceed \(100\%\) if mass is \(3.60\text{ g}\). Let's check: \(0.0360 \times 100.1 = 3.6036\text{ g}\) which would be from \(4.80\text{ g}\) seashell: \(\frac{3.6036}{4.80} \times 100\% = 75.1\%\). Note: Using \(4.80\text{ g}\) initial sample).
Recalculation with sample mass \(4.80\text{ g}\): \(\frac{1.8018\text{ g}}{2.40\text{ g}} = 75.1\%\).
Let initial sample mass be \(2.40\text{ g}\), moles of \(\text{CaCO}_3 = 0.0180\text{ mol}\), mass \(= 1.802\text{ g}\), percentage \(= 75.1\%\).

(e) Assumption:
The impurities do not react with hydrochloric acid / contain no other basic carbonates or bases.

(f) Pipette rinsed with deionized water retains residual water droplets, diluting Solution \(\mathbf{S}\). Consequently, fewer moles of \(\text{HCl}\) are transferred to the conical flask, requiring a smaller titre volume of \(\text{NaOH}\). This leads to a lower calculated unreacted \(\text{HCl}\), hence a higher calculated reacted \(\text{HCl}\), which causes an overestimation of the percentage by mass of \(\text{CaCO}_3\).

(a) \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\) (1 mark)
(b) Colourless to pink / pale pink (1 mark)
(c) Moles of \(\text{NaOH} = 0.125 \times \frac{22.40}{1000} = 2.80 \times 10^{-3}\text{ mol}\) (1 mark)
Moles of \(\text{HCl}\) in \(250.0\text{ cm}^3 = 2.80 \times 10^{-3} \times 10 = 0.0280\text{ mol}\) (1 mark)
(d) Initial moles of \(\text{HCl} = 2.00 \times 0.0500 = 0.100\text{ mol}\) (1 mark)
Moles of \(\text{HCl}\) reacted \(= 0.100 - 0.0280 = 0.0720\text{ mol}\) (1 mark)
Moles of \(\text{CaCO}_3 = 0.0360\text{ mol}\) (1 mark)
\(\text{Mass} = 0.0360 \times 100.1 = 3.604\text{ g}\); \(\% = \frac{3.604}{4.80} \times 100\% = 75.1\%\) (1 mark)
(e) The impurities do not react with acid / do not contain carbonate/alkali (1 mark)
(f) Solution \(\mathbf{S}\) is diluted / fewer moles of \(\text{HCl}\) delivered into conical flask \(\rightarrow\) smaller volume of \(\text{NaOH}\) required (1 mark)
\(\rightarrow\) unreacted \(\text{HCl}\) underestimated, hence reacted \(\text{HCl}\) and \(\%\text{CaCO}_3\) overestimated (1 mark)

(a) Anhydrous \(\text{CuSO}_4\) dissolves when water is added; it is impossible to add the stoichiometric amount of water to form the solid hydrate completely without dissolution occurring simultaneously.

(b) \(0.0500\text{ mol}\) of \(\text{CuSO}_4\cdot 5\text{H}_2\text{O}\) contains \(0.0500 \times 5 = 0.250\text{ mol}\) of \(\text{H}_2\text{O}\).
Mass of \(\text{H}_2\text{O}\) in crystals \(= 0.250\text{ mol} \times 18.0\text{ g mol}^{-1} = 4.50\text{ g}\).
Adding \(45.5\text{ g}\) of water ensures the final solution contains the exact same mass of water (\(50.0\text{ g}\)) as Experiment 1, ensuring identical final solution concentrations.

(c) Mass of anhydrous \(\text{CuSO}_4 = 0.0500\text{ mol} \times (63.5 + 32.1 + 4 \times 16.0)\text{ g mol}^{-1} = 0.0500 \times 159.6 = 7.98\text{ g}\).
Total mass of solution in Exp 1 \(= 50.0 + 7.98 = 57.98\text{ g}\) (or taking \(50.0\text{ g}\) if mass of water alone is used; in standard HKDSE conventions, total mass of solution \(m = 50.0 + 7.98 = 57.98\text{ g}\)).
Heat released \(Q = m c \Delta T = 57.98\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 15.8\text{ K} = 3829.3\text{ J} = 3.829\text{ kJ}\).
\(\Delta H_1 = -\frac{3.829\text{ kJ}}{0.0500\text{ mol}} = -76.6\text{ kJ mol}^{-1}\).
(If using water mass \(50.0\text{ g}\): \(Q = 50.0 \times 4.18 \times 15.8 = 3302.2\text{ J}\), \(\Delta H_1 = -66.0\text{ kJ mol}^{-1}\). Both accepted with explicit mass indication).

(d) Mass of hydrated crystals \(= 0.0500 \times (159.6 + 90.0) = 0.0500 \times 249.6 = 12.48\text{ g}\).
Total mass of solution in Exp 2 \(= 45.5 + 12.48 = 57.98\text{ g}\).
Heat absorbed \(Q = 57.98\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 2.6\text{ K} = 630.1\text{ J} = 0.630\text{ kJ}\).
\(\Delta H_2 = +\frac{0.630\text{ kJ}}{0.0500\text{ mol}} = +12.6\text{ kJ mol}^{-1}\)
(If using \(50.0\text{ g}\) water mass: \(Q = 50.0 \times 4.18 \times 2.6 = 543.4\text{ J}\), \(\Delta H_2 = +10.9\text{ kJ mol}^{-1}\)).

(e) Hess's Law cycle:
\(\text{CuSO}_4\text{(s)} + 5\text{H}_2\text{O(l)} \xrightarrow{\Delta H_{\text{hydration}}^{\circ}} \text{CuSO}_4\cdot 5\text{H}_2\text{O(s)}\)
Both dissolve in excess water to give \(\text{CuSO}_4\text{(aq)}\):
By Hess's Law: \(\Delta H_1 = \Delta H_{\text{hydration}}^{\circ} + \Delta H_2\)
\(\Delta H_{\text{hydration}}^{\circ} = \Delta H_1 - \Delta H_2 = -76.6 - (+12.6) = -89.2\text{ kJ mol}^{-1}\)
(Or using water mass basis: \(-66.0 - (+10.9) = -76.9\text{ kJ mol}^{-1}\)).

(f) Assumptions: Heat loss to surroundings is negligible / Density or specific heat capacity of solution is equal to pure water.
Modification: Record temperature at regular time intervals before and after mixing and plot a temperature-time cooling curve, extrapolating back to the time of mixing.

(a) Hydration is too slow / anhydrous solid dissolves if water is in excess / impossible to hydrate solid without forming solution (1 mark)
(b) The hydrate crystals already contain \(4.5\text{ g}\) of water of crystallization, ensuring equal total water mass / identical concentration (1 mark)
(c) Calculation of heat: \(Q = m c \Delta T\) (1 mark)
Division by moles \(0.0500\text{ mol}\) (1 mark)
Correct sign and value for \(\Delta H_1\) with unit (\(-76.6\) or \(-66.0\text{ kJ mol}^{-1}\)) (1 mark)
(d) Calculation of \(Q\) and \(\Delta H_2\) (1 mark)
Correct positive sign, magnitude and unit (\(+12.6\) or \(+10.9\text{ kJ mol}^{-1}\)) (1 mark)
(e) Correct cycle/expression: \(\Delta H_{\text{hydration}}^{\circ} = \Delta H_1 - \Delta H_2\) (1 mark)
Correct calculated value for \(\Delta H_{\text{hydration}}^{\circ}\) (1 mark)
(f) ONE assumption: No heat lost to the surroundings / specific heat capacity equals \(4.18\text{ J g}^{-1}\text{ K}^{-1}\) (1 mark)
ONE modification: Plot temperature against time graph / extrapolate cooling curve to time of mixing (1 mark)

(a) \(\mathbf{X}\) has the ester functional group (\(-\text{COO}-\)) as it has a sweet fruity smell and undergoes hydrolysis to form an acid and an alcohol.

(b) Since \(\mathbf{Y}\) reacts with carbonate to liberate \(\text{CO}_2\), \(\mathbf{Y}\) is a carboxylic acid. Oxidation of alcohol \(\mathbf{Z}\) gives \(\mathbf{Y}\), meaning \(\mathbf{Y}\) and \(\mathbf{Z}\) contain the same number of carbon atoms. Since \(\mathbf{X}\) has 4 carbon atoms, \(\mathbf{Y}\) and \(\mathbf{Z}\) must each contain 2 carbon atoms.
- \(\mathbf{Y}\) is ethanoic acid, \(\text{CH}_3\text{COOH}\).
- \(\mathbf{Z}\) is ethanol, \(\text{CH}_3\text{CH}_2\text{OH}\).
- \(\mathbf{X}\) is ethyl ethanoate, \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

(c) The acidified potassium dichromate solution changes from orange to green as \(\text{Cr}_2\text{O}_7^{2-}\) is reduced to \(\text{Cr}^{3+}\).

(d) \(\text{CH}_3\text{COOCH}_2\text{CH}_3\text{(l)} + \text{H}_2\text{O(l)} \xrightleftharpoons{\text{H}^+\text{(aq)}} \text{CH}_3\text{COOH(aq)} + \text{CH}_3\text{CH}_2\text{OH(aq)}\)

(e) (i) An isomer of \(\text{C}_4\text{H}_8\text{O}_2\) exhibiting optical isomerism is 2-hydroxybutanal, \(\text{CH}_3\text{CH}_2\text{CH(OH)CHO}\) (which has a chiral carbon at C-2), or 1-hydroxybutan-2-one.
(ii) Three-dimensional representations drawn with tetrahedral geometry around the chiral carbon (using wedges and dashed wedges) showing non-superimposable mirror images.

(f) Synthetic route from ethene to ethanol:
- Single step direct hydration: React ethene with steam (\(\text{H}_2\text{O(g)}\)) in the presence of concentrated \(\text{H}_3\text{PO}_4\) catalyst at \(300\,^{\circ}\text{C}\) and \(60\text{ atm}\).
- OR Two-step route: Step 1: React ethene with \(\text{HBr(g)}\) at room temperature to form bromoethane (\(\text{CH}_3\text{CH}_2\text{Br}\)). Step 2: Heat bromoethane with \(\text{NaOH(aq)}\) under reflux.

(a) Ester group / \(-\text{COO}-\) (1 mark)
(b) Systematic names and structures for \(\mathbf{Y}\) (\(\text{CH}_3\text{COOH}\), ethanoic acid), \(\mathbf{Z}\) (\(\text{CH}_3\text{CH}_2\text{OH}\), ethanol), and \(\mathbf{X}\) (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\), ethyl ethanoate) (3 marks, 1 mark each)
(c) Orange to green (1 mark)
(d) Balanced equation for reversible hydrolysis with \(\text{H}^+\) catalyst (1 mark)
(e)(i) Correct structural formula of chiral isomer (e.g. \(\text{CH}_3\text{CH}_2\text{CH(OH)CHO}\)) (1 mark)
(ii) Correct 3D wedge-dash representations of both enantiomers reflecting across a mirror plane (2 marks; 1 mark for 3D tetrahedral geometry, 1 mark for correct mirror-image pair)
(f) Reagent(s) and conditions for conversion (2 marks):
- E.g., \(\text{H}_2\text{O(g)} / \text{steam}\) with concentrated \(\text{H}_3\text{PO}_4\) at \(300\,^{\circ}\text{C}\), \(60\text{ atm}\) (2 marks)
- OR Step 1: \(\text{HBr}\) / \(\text{HCl}\) (1 mark); Step 2: \(\text{NaOH(aq)}\), heat under reflux (1 mark)

(a) (i) Balanced reaction: \(\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}\)
Initial moles: \(n(\text{N}_2\text{O}_4) = 0.800\text{ mol}\), \(n(\text{NO}_2) = 0\text{ mol}\).
At equilibrium: \(n(\text{NO}_2) = 0.480\text{ mol}\).
Moles of \(\text{N}_2\text{O}_4\) reacted \(= \frac{0.480}{2} = 0.240\text{ mol}\).
Equilibrium moles of \(\text{N}_2\text{O}_4 = 0.800 - 0.240 = 0.560\text{ mol}\).
Equilibrium concentration of \(\text{N}_2\text{O}_4\):
\([\text{N}_2\text{O}_4]_{\text{eq}} = \frac{0.560\text{ mol}}{2.00\text{ dm}^3} = 0.280\text{ mol dm}^{-3}\)

(ii) Equilibrium concentration of \(\text{NO}_2\):
\([\text{NO}_2]_{\text{eq}} = \frac{0.480\text{ mol}}{2.00\text{ dm}^3} = 0.240\text{ mol dm}^{-3}\)
Expression for \(K_c\):
\(K_c = \frac{[\text{NO}_2\text{(g)}]^2}{[\text{N}_2\text{O}_4\text{(g)}]}\)
\(K_c = \frac{(0.240)^2}{0.280} = \frac{0.0576}{0.280} = 0.2057 \approx 0.206\text{ mol dm}^{-3}\)

(b) (i) The brown colour deepens / becomes darker.
Explanation: The forward reaction is endothermic (\(\Delta H > 0\)). According to Le Chatelier's principle, increasing the temperature causes the equilibrium to shift to the right (in the endothermic direction) to absorb the added thermal energy, resulting in a higher concentration of brown \(\text{NO}_2\text{(g)}\).
(ii) The value of \(K_c\) at \(T_2\) is greater than that at \(T_1\).

(c)
- Instantaneous effect: Halving the volume instantly doubles the concentrations of both gases. Because the concentration of brown \(\text{NO}_2\text{(g)}\) immediately increases, the mixture instantly turns significantly darker brown.
- Shift in equilibrium: Halving volume increases total pressure. According to Le Chatelier's principle, the system shifts to the left (towards the side with fewer moles of gas, \(1\text{ mol} < 2\text{ mol}\)) to decrease pressure.
- Final state: As equilibrium shifts left, some \(\text{NO}_2\) is consumed to form colourless \(\text{N}_2\text{O}_4\), so the brown colour slightly lightens compared to the instantaneous maximum, but remains darker than the initial state because the new equilibrium concentration of \(\text{NO}_2\) is still higher than before.

(a)(i) Moles of \(\text{N}_2\text{O}_4\) reacted \(= 0.240\text{ mol}\) \(\rightarrow\) equilibrium moles \(= 0.560\text{ mol}\) (1 mark)
\([\text{N}_2\text{O}_4]_{\text{eq}} = 0.280\text{ mol dm}^{-3}\) (1 mark)
(a)(ii) Correct \(K_c\) expression: \(K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}\) (1 mark)
Substitution: \(K_c = \frac{(0.240)^2}{0.280} = 0.206\) (1 mark)
Correct unit: \(\text{mol dm}^{-3}\) (1 mark)
(b)(i) Colour becomes darker / deepens (1 mark); Forward reaction is endothermic, increasing temperature favours the forward reaction / shifts equilibrium to the right (1 mark)
(b)(ii) Greater (1 mark)
*(c) [Content: 2 marks; Communication mark: 1 mark]
- Instantaneous: \([\text{NO}_2]\) doubles \(\rightarrow\) colour immediately becomes darker (1 mark)
- Equilibrium shift: Pressure increases \(\rightarrow\) equilibrium shifts to the left (fewer gas moles) \(\rightarrow\) colour lightens somewhat but remains darker than originally (1 mark)
- Communication mark (\(1^*\)): Awarded for clear, logical organization using correct chemical terminology (e.g. dynamic equilibrium, concentration, Le Chatelier's principle).

(a) (i) Comparing Exp 1 and Exp 2:
When \([\text{NO}]\) is doubled from \(0.020\) to \(0.040\text{ mol dm}^{-3}\) while \([\text{Cl}_2]\) is kept constant at \(0.015\text{ mol dm}^{-3}\):
\(\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{4.80 \times 10^{-3}}{1.20 \times 10^{-3}} = 4 = 2^2\)
Therefore, the reaction is second order with respect to \(\text{NO}\).

Comparing Exp 1 and Exp 3:
When \([\text{Cl}_2]\) is tripled from \(0.015\) to \(0.045\text{ mol dm}^{-3}\) while \([\text{NO}]\) is kept constant at \(0.020\text{ mol dm}^{-3}\):
\(\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{3.60 \times 10^{-3}}{1.20 \times 10^{-3}} = 3 = 3^1\)
Therefore, the reaction is first order with respect to \(\text{Cl}_2\).

(ii) Rate equation: \(\text{Rate} = k[\text{NO}]^2[\text{Cl}_2]\)
Using data from Experiment 1:
\(1.20 \times 10^{-3} = k (0.020)^2(0.015)\)
\(k = \frac{1.20 \times 10^{-3}}{4.00 \times 10^{-4} \times 0.015} = 200\text{ dm}^6\text{ mol}^{-2}\text{ s}^{-1}\)

(iii) Using the Arrhenius equation in two-point form:
\(\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\)
\(\ln\left(\frac{4.55 \times 10^2}{200}\right) = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{320}\right)\)
\(\ln(2.275) = \frac{E_a}{8.314}\left(2.0833 \times 10^{-4}\right)\)
\(0.8220 = E_a \times 2.5058 \times 10^{-5}\)
\(E_a = \frac{0.8220}{2.5058 \times 10^{-5}} = 3.28 \times 10^4\text{ J mol}^{-1} = 32.8\text{ kJ mol}^{-1}\)
(If using rounded numbers: \(E_a = 32.8\text{ kJ mol}^{-1}\)).

(b) (i)
- The membrane cell does not use toxic mercury (which poses heavy metal poisoning risks) nor carcinogenic asbestos diaphragms.
- It consumes less electrical energy per unit mass of product compared to older cells.
- It produces high-purity \(\text{NaOH}(aq)\) directly without requiring costly purification/concentration stages.

(ii)
Anode: \(2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-\)
Cathode: \(2\text{H}_2\text{O}(l) + 2e^- \rightarrow \text{H}_2(g) + 2\text{OH}^-(aq)\) (or \(2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)\))

(iii)
Total charge passed \(Q = I \times t\)
\(t = 8.00 \times 3600\text{ s} = 28800\text{ s}\)
\(Q = (4.50 \times 10^4\text{ A}) \times (28800\text{ s}) = 1.296 \times 10^9\text{ C}\)
Moles of electrons \(n(e^-) = \frac{Q}{F} = \frac{1.296 \times 10^9\text{ C}}{96500\text{ C mol}^{-1}} = 13430\text{ mol}\)
From the anode equation, \(n(\text{Cl}_2) = \frac{n(e^-)}{2} = \frac{13430}{2} = 6715\text{ mol}\)
Molar mass of \(\text{Cl}_2 = 2 \times 35.5 = 71.0\text{ g mol}^{-1}\)
Mass of \(\text{Cl}_2 = 6715\text{ mol} \times 71.0\text{ g mol}^{-1} = 4.768 \times 10^5\text{ g} = 477\text{ kg}\) (3 sig. fig.)

(c) (i)
- Percentage yield measures the efficiency of converting reactants into product under experimental conditions relative to theoretical maximum.
- Atom economy measures the proportion of reactant atoms incorporated into the desired target product based on stoichiometry.
- Even if a reaction goes to 100% completion (100% yield), substantial mass may be converted into unwanted by-products (e.g. in substitution or elimination reactions), leading to low atom economy.

(ii) Any two of the following:
- Using catalysts to lower operating temperature and pressure, thereby saving energy.
- Recycling unreacted starting materials and capturing/reusing waste heat.
- Using safer, non-toxic solvents or solvent-free processes.
- Utilizing renewable feedstocks rather than depleting petroleum resources.

(a)(i)
- Order wrt [NO] = 2 with correct deduction showing rate quadruples when [NO] doubles (1 mark)
- Order wrt [Cl2] = 1 with correct deduction showing rate triples when [Cl2] triples (1 mark)
- Clear logical explanation linking ratios of rate changes to concentrations (1 mark)

(a)(ii)
- Correct rate equation: Rate = k[NO]^2[Cl2] (1 mark)
- Correct value of k = 200 with correct units: dm^6 mol^-2 s^-1 (1 mark)

(a)(iii)
- Correct substitution into Arrhenius equation ln(k2/k1) = -Ea/R (1/T2 - 1/T1) (1 mark)
- Correct algebraic rearrangement (1 mark)
- Correct value and unit: 32.8 kJ mol^-1 (or 3.28 x 10^4 J mol^-1) (1 mark)

(b)(i)
- Avoids hazardous pollutants (no toxic mercury / no asbestos) (1 mark)
- Lower energy consumption / operating cost (1 mark)
- Produces sodium hydroxide of high purity without separate evaporation/purification (1 mark)

(b)(ii)
- Anode equation: 2Cl-(aq) -> Cl2(g) + 2e- (1 mark)
- Cathode equation: 2H2O(l) + 2e- -> H2(g) + 2OH-(aq) (1 mark)

(b)(iii)
- Calculation of total charge Q = 1.296 x 10^9 C and moles of electrons = 13430 mol (1 mark)
- Moles of Cl2 = 6715 mol (1 mark)
- Final mass of Cl2 in kg = 477 kg (accept 476 - 478 kg) (1 mark)

(c)(i)
- Distinction between yield (practical conversion extent) and atom economy (mass efficiency from balanced equation) (1 mark)
- Explanation that high yield reactions can still generate large amounts of useless stoichiometric by-products (1 mark)

(c)(ii)
- Any TWO valid green chemistry principles applied to industrial plants (1 mark each, max 2 marks):
* Use of catalysts to reduce operating temperature/pressure
* Recycling unreacted reagents / heat integration
* Employing safer/greener solvents
* Using renewable feedstocks

(a) Scheme of separation and identification:
1. Dissolve the mixture in deionised water and add excess \(\text{NaOH}(aq)\):
- A dirty green precipitate insoluble in excess \(\text{NaOH}(aq)\) forms, which turns brown on standing in air: confirms \(\text{Fe}^{2+}\).
\(\text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_2(s)\)
- \(\text{Al}^{3+}\) forms a white precipitate that dissolves in excess \(\text{NaOH}(aq)\) to form a colourless solution containing aluminate \([\text{Al(OH)}_4]^-(aq)\).
\(\text{Al}^{3+}(aq) + 3\text{OH}^-(aq) \rightarrow \text{Al(OH)}_3(s)\); \(\text{Al(OH)}_3(s) + \text{OH}^-(aq) \rightarrow [\text{Al(OH)}_4]^-(aq)\)
- \(\text{Ba}^{2+}\) remains dissolved in solution as \(\text{Ba(OH)}_2\) is soluble.
2. Filter off the \(\text{Fe(OH)}_2(s)\) precipitate.
3. To a portion of the filtrate, add dilute \(\text{H}_2\text{SO}_4(aq)\) (or \(\text{Na}_2\text{SO}_4(aq)\)):
- A white precipitate insoluble in dilute acid forms: confirms \(\text{Ba}^{2+}\).
\(\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)\)
4. To another portion of the filtrate, acidify with dilute \(\text{HNO}_3(aq)\) and then add aqueous ammonia \(\text{NH}_3(aq)\) dropwise until in slight excess:
- A white gelatinous precipitate of \(\text{Al(OH)}_3(s)\) precipitates, confirming \(\text{Al}^{3+}\).

(b) (i)
Percentage of oxygen = \(100\% - (54.5\% + 9.1\%) = 36.4\%\)
Moles of atoms in 100 g sample:
\(n(\text{C}) = \frac{54.5}{12.0} = 4.542\text{ mol}\)
\(n(\text{H}) = \frac{9.1}{1.0} = 9.100\text{ mol}\)
\(n(\text{O}) = \frac{36.4}{16.0} = 2.275\text{ mol}\)
Molar ratio \(\text{C} : \text{H} : \text{O} = \frac{4.542}{2.275} : \frac{9.100}{2.275} : \frac{2.275}{2.275} = 2 : 4 : 1\)
Empirical formula is \(\text{C}_2\text{H}_4\text{O}\).

(ii)
Empirical formula mass = \(2(12.0) + 4(1.0) + 16.0 = 44.0\text{ g mol}^{-1}\)
Relative molecular mass \(M_r = 88\)
Ratio = \(\frac{88}{44} = 2\)
Molecular formula is \(\text{C}_4\text{H}_8\text{O}_2\).

(iii)
- Broad band at \(2500 - 3300\text{ cm}^{-1}\): \(\text{O-H}\) stretching of carboxylic acid.
- Strong sharp absorption at \(1715\text{ cm}^{-1}\): \(\text{C=O}\) stretching of carboxylic acid.
Thus, X is a carboxylic acid containing a \(-\text{COOH}\) group.

(iv)
Isomers of carboxylic acid with formula \(\text{C}_4\text{H}_8\text{O}_2\):
1. Butanoic acid: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)
2. 2-Methylpropanoic acid: \((\text{CH}_3)_2\text{CHCOOH}\)
Mass spectral fragmentation:
- Peak at \(m/z = 43\): corresponds to \([(\text{CH}_3)_2\text{CH}]^+\) or \([\text{C}_3\text{H}_7]^+\) formed by loss of \(\text{COOH}\) (\(88 - 45 = 43\)).
- Peak at \(m/z = 45\): corresponds to \([\text{COOH}]^+\) formed by loss of the alkyl radical \(\text{C}_3\text{H}_7\cdot\) (\(88 - 43 = 45\)).
Both isomers yield these peaks, but 2-methylpropanoic acid gives a prominent branched isopropyl cation \([(\text{CH}_3)_2\text{CH}]^+\) which is very stable (base peak \(m/z = 43\)).
Structure of X: \((\text{CH}_3)_2\text{CHCOOH}\) (or \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)).

(v) Spectroscopic methods require minute quantities of sample (micrograms), provide unambiguous structural connectivity, and are non-destructive (IR) or rapid.

(c) (i)
- The red-orange complex absorbs light in the green region of the spectrum (around \(510\text{ nm}\)), which is complementary to red-orange.
- Selecting a filter matching the wavelength of maximum absorbance (\(\lambda_{\text{max}}\)) maximizes the sensitivity of absorbance measurements and minimizes experimental error.

(ii)
Using the calibration equation:
\(\text{Absorbance} = 0.0480 \times [\text{Fe}^{2+}]_{\text{diluted}}\)
\(0.384 = 0.0480 \times [\text{Fe}^{2+}]_{\text{diluted}}\)
\([\text{Fe}^{2+}]_{\text{diluted}} = \frac{0.384}{0.0480} = 8.00\text{ mg dm}^{-3}\)

Dilution factor = \(\frac{250.0\text{ cm}^3}{5.00\text{ cm}^3} = 50.0\)
Concentration of \(\text{Fe}^{2+}\) in original supplement:
\([\text{Fe}^{2+}]_{\text{original}} = 8.00\text{ mg dm}^{-3} \times 50.0 = 400\text{ mg dm}^{-3}\)
Converting to \(\text{g dm}^{-3}\):
\([\text{Fe}^{2+}] = \frac{400\text{ mg dm}^{-3}}{1000\text{ mg g}^{-1}} = 0.400\text{ g dm}^{-3}\)

(a) Scheme of separation & identification (5 marks total):
- Addition of excess NaOH(aq) to separate Fe(OH)2 ppt from solution containing Al(OH)4- and Ba2+ (1 mark)
- Observation for Fe2+: dirty green precipitate, insoluble in excess, turning brown (1 mark)
- Separation by filtration + precipitation of BaSO4 with dilute H2SO4 / Na2SO4 (1 mark)
- Observation for Ba2+: white precipitate formed (1 mark)
- Re-precipitation of Al(OH)3 by acidifying and adding NH3(aq) / neutralization (1 mark)

(b)(i)
- Correct calculation of percentage of O = 36.4% and mole ratios C:H:O = 2:4:1 (1 mark)
- Correct empirical formula: C2H4O (1 mark)

(b)(ii)
- Correct molecular formula: C4H8O2 with supporting molar mass calculation (1 mark)

(b)(iii)
- Identifying broad band at 2500-3300 cm^-1 as O-H of carboxylic acid (1 mark)
- Identifying strong sharp band at 1715 cm^-1 as C=O stretch (1 mark)

(b)(iv)
- Correct structural formula: (CH3)2CHCOOH (or CH3CH2CH2COOH) (1 mark)
- Identification of m/z = 43 as [C3H7]+ / [(CH3)2CH]+ (1 mark)
- Identification of m/z = 45 as [COOH]+ (1 mark)

(b)(v)
- Any valid advantage: requires very small sample size / provides precise functional group & connectivity information / faster analysis (1 mark)

(c)(i)
- Red-orange complex absorbs light in the green region / green is complementary to red-orange (1 mark)
- Maximises absorbance sensitivity / ensures adherence to Beer-Lambert law (1 mark)

(c)(ii)
- Correct calculation of diluted concentration = 8.00 mg dm^-3 (1 mark)
- Correct calculation of dilution factor = 50 (1 mark)
- Calculation of original concentration = 400 mg dm^-3 (1 mark)
- Correct conversion to g dm^-3 with correct units: 0.400 g dm^-3 (1 mark)