2025 Cambridge IGCSE Chemistry (0620) 模擬試題連答案詳解

1. Calculate the number of moles of \(\text{Cu(NO}_3)_2\):
\(\text{Moles} = \frac{5.64\text{ g}}{188\text{ g/mol}} = 0.030\text{ mol}\).

2. Determine the molar ratio of reactant to total gas products from the balanced chemical equation:
\(2\text{ moles of Cu(NO}_3)_2\) produce \(4\text{ moles of NO}_2\) and \(1\text{ mole of O}_2\), giving a total of \(5\text{ moles}\) of gas.
This means the ratio of reactant to total gas is \(2 : 5\).

3. Calculate the moles of gas produced:
\(\text{Moles of gas} = 0.030\text{ mol} \times \frac{5}{2} = 0.075\text{ mol}\).

4. Calculate the volume of gas at r.t.p.:
\(\text{Volume} = 0.075\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 1.80\text{ dm}^3\).

[1 mark] Option A is correct.
- Award 1 mark for the correct calculation of moles of copper(II) nitrate (0.030 mol), converting this to the total moles of gas (0.075 mol), and multiplying by 24.0 to obtain 1.80 dm³.

During the electrolysis of concentrated aqueous sodium chloride:
- Chloride ions, \(\text{Cl}^-\), migrate to the anode (positive electrode) and are oxidised to form chlorine gas, \(\text{Cl}_2\).
- Hydrogen ions, \(\text{H}^+\), from water migrate to the cathode (negative electrode) and are preferentially reduced (since hydrogen is less reactive than sodium) to form hydrogen gas, \(\text{H}_2\).
- This leaves sodium ions, \(\text{Na}^+\), and hydroxide ions, \(\text{OH}^-\), in the solution, forming sodium hydroxide, which is alkaline. Consequently, the pH of the remaining electrolyte increases.

[1 mark] Option A is correct. Award 1 mark for correctly identifying that chlorine is produced at the anode, hydrogen is produced at the cathode, and the pH of the remaining electrolyte increases.

1. Calculate energy required to break bonds (endothermic process):
- \(1 \times \text{N}\equiv\text{N} = 945\text{ kJ/mol}\)
- \(3 \times \text{H}-\text{H} = 3 \times 436 = 1308\text{ kJ/mol}\)
Total energy absorbed \(= 945 + 1308 = 2253\text{ kJ/mol}\).

2. Calculate energy released when bonds are formed (exothermic process):
- Ammonia, \(\text{NH}_3\), contains three \(\text{N}-\text{H}\) bonds. For \(2\text{NH}_3\), there are \(6\text{ N}-\text{H}\) bonds.
- \(6 \times 391 = 2346\text{ kJ/mol}\).
Total energy released \(= 2346\text{ kJ/mol}\).

3. Calculate the overall energy change:
\(\text{Energy change} = \text{Energy absorbed} - \text{Energy released} = 2253 - 2346 = -93\text{ kJ/mol}\).

[1 mark] Option A is correct.
- Award 1 mark for calculating the total energy required to break reactants' bonds (+2253 kJ/mol), the total energy released when products' bonds form (-2346 kJ/mol), and determining the final enthalpy change as -93 kJ/mol.

- A steeper initial gradient means the rate of reaction is faster.
- Leveling off at the same final volume means the total moles of limiting reactant (which is the hydrochloric acid, since calcium carbonate is in excess) remains unchanged.
- Using powdered calcium carbonate increases its surface area, which increases the frequency of successful collisions and thus the rate of reaction (making the curve steeper) but does not change the amount of limiting reactant, so the final volume of carbon dioxide remains the same.
- Options B and C would increase the moles of hydrochloric acid, which is the limiting reactant, thereby increasing the total volume of gas produced.
- Option D would decrease the temperature of the acid, which decreases the rate of reaction, making the curve less steep.

[1 mark] Option A is correct.
- Award 1 mark for identifying that increasing the surface area of the solid reactant increases the rate of reaction (steeper curve) without changing the final yield of gas.

- Temperature: The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium in the exothermic direction (to the right) to release heat, thus increasing the yield of \(\text{SO}_3\). Therefore, a low temperature is required.
- Pressure: The forward reaction involves a decrease in the number of gaseous moles (from 3 moles on the left to 2 moles on the right). Increasing the pressure shifts the equilibrium to the side with fewer gas molecules (to the right) to reduce pressure, increasing the yield of \(\text{SO}_3\). Therefore, a high pressure is required.
- Catalyst: A catalyst increases the rates of both the forward and reverse reactions equally. It speeds up the rate at which equilibrium is reached, but has no effect on the position of the equilibrium or the yield.

[1 mark] Option A is correct.
- Award 1 mark for correctly identifying that a low temperature shifts the equilibrium to the right, a high pressure shifts the equilibrium to the right, and a catalyst has no effect on the position of equilibrium.

- To prepare an insoluble salt like lead(II) sulfate, a precipitation reaction is used. This requires mixing two soluble starting materials.
- Lead(II) nitrate and sodium sulfate are both soluble in water and will react to form the insoluble precipitate lead(II) sulfate.
- The correct experimental steps to isolate a pure, dry sample of an insoluble precipitate are:
1. Filter the reaction mixture to collect the precipitate as the residue on the filter paper.
2. Wash the residue with distilled water to remove any soluble impurities.
3. Dry the residue (precipitate) in a warm oven or between sheets of filter paper.
- Evaporation/crystallisation is used to retrieve soluble salts from a solution, so options B and D are incorrect. Solid lead(II) oxide and solid lead(II) carbonate do not react completely with sulfuric acid because a layer of insoluble lead(II) sulfate quickly coats the solids, stopping the reaction (option C is incorrect).

[1 mark] Option A is correct.
- Award 1 mark for identifying that two soluble starting materials are required to make an insoluble salt by precipitation, and that the precipitate must be filtered, washed, and dried.

- Reaction with aqueous sodium carbonate to produce carbon dioxide gas is the characteristic test for a carboxylic acid.
- The molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) fits the general formula for both carboxylic acids and esters (\(\text{C}_n\text{H}_{2n}\text{O}_2\)).
- Among the options:
- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\) (butanoic acid) is a carboxylic acid with 4 carbon atoms, 8 hydrogen atoms, and 2 oxygen atoms. It reacts with sodium carbonate to produce \(\text{CO}_2\).
- \(\text{CH}_3\text{COOCH}_2\text{CH}_3\) (ethyl ethanoate) is an ester; it has the same molecular formula but does not react with sodium carbonate.
- \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\) (butan-1-ol) is an alcohol and does not react with sodium carbonate.
- \(\text{CH}_3\text{CH}_2\text{COCH}_3\) (butanone) is a ketone and does not react with sodium carbonate.

[1 mark] Option A is correct.
- Award 1 mark for identifying that compound X must be a carboxylic acid due to its reaction with sodium carbonate, and verifying that butanoic acid has the molecular formula C₄H₈O₂.

- Test 1: Both \(\text{Fe}^{2+}\) and \(\text{Cr}^{3+}\) produce green precipitates with aqueous sodium hydroxide. However, the chromium(III) hydroxide precipitate is soluble in excess sodium hydroxide, forming a green solution, while the iron(II) hydroxide precipitate is insoluble in excess. Therefore, the cation present is \(\text{Fe}^{2+}\).
- Test 2: The addition of dilute nitric acid followed by aqueous barium nitrate is the test for sulfate ions, \(\text{SO}_4^{2-}\). The formation of a white precipitate (barium sulfate) confirms the presence of sulfate ions.
- Therefore, the solution contains \(\text{Fe}^{2+}\) and \(\text{SO}_4^{2-}\).

[1 mark] Option A is correct.
- Award 1 mark for identifying that Fe²⁺ produces a green precipitate insoluble in excess NaOH, and that the sulfate test yields a white precipitate with barium nitrate.

First, calculate the molar mass of sodium carbonate, \(\text{Na}_2\text{CO}_3\): \(M_r = (2 \times 23) + 12 + (3 \times 16) = 106\text{ g/mol}\). Next, find the number of moles of sodium carbonate used: \(\text{moles} = 5.3\text{ g} / 106\text{ g/mol} = 0.05\text{ mol}\). The chemical equation for the reaction is: \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\). The stoichiometry shows a 1:1 ratio between \(\text{Na}_2\text{CO}_3\) and \(\text{CO}_2\). Therefore, \(0.05\text{ mol}\) of \(\text{CO}_2\) gas is produced. Finally, calculate the volume of gas at r.t.p.: \(\text{Volume} = 0.05\text{ mol} \times 24\text{ dm}^3\text{/mol} = 1.2\text{ dm}^3\).

Award 1 mark for the correct option A. Deduct/reject other options as they represent incorrect stoichiometric calculations or incorrect molar volume scaling.

In this displacement reaction, the bromide ions (\(\text{Br}^-\)) lose electrons to form bromine molecules (\(\text{Br}_2\)). The loss of electrons is oxidation, so bromide ions are oxidized. Chlorine molecules (\(\text{Cl}_2\)) gain electrons to form chloride ions (\(\text{Cl}^-\)), so chlorine is reduced. Potassium ions (\(\text{K}^+\)) are spectator ions and do not undergo redox.

Award 1 mark for the correct option A. Reject B, C, and D because they contain incorrect definitions of redox, oxidation states, or electron transfer.

Increasing the temperature increases the average kinetic energy of the reacting particles. Consequently, a greater fraction of the colliding particles possess energy equal to or greater than the activation energy (\(E_a\)), leading to a higher frequency of successful collisions.

Award 1 mark for the correct option C. Reject A (catalysts decrease activation energy), B (concentration/pressure changes particle distance), and D (particles move faster at higher temperatures).

A methanol molecule has three \(\text{C-H}\) bonds, one \(\text{C-O}\) bond, and one \(\text{O-H}\) bond, making a total of 5 single covalent bonds (5 bonding pairs). The carbon atom uses all 4 of its outer-shell electrons for bonding. Each hydrogen uses its single electron. The oxygen atom has 6 outer-shell electrons, uses 2 for bonding, and has 4 non-bonding outer electrons remaining, which form 2 non-bonding pairs (lone pairs).

Award 1 mark for the correct option A. Reject B, C, and D due to incorrect counts of shared or unshared electron pairs.

The formation of a green precipitate with sodium hydroxide that is insoluble in excess indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). The formation of a white precipitate with acidified barium nitrate indicates the presence of sulfate ions, \(\text{SO}_4^{2-}\). Therefore, the salt is iron(II) sulfate.

Award 1 mark for the correct option A. Reject B (iron(III) forms a red-brown precipitate), C (chloride ions would form a precipitate with silver nitrate, not barium nitrate), and D (copper(II) forms a light blue precipitate).

The forward reaction is endothermic (\(\Delta H > 0\)), so the reverse reaction is exothermic. Decreasing the temperature shifts the equilibrium in the exothermic direction (to the left), increasing the concentration of colorless \(\text{N}_2\text{O}_4\) and making the mixture lighter brown. Increasing the pressure shifts the equilibrium to the side with fewer moles of gas (the left side has 1 mole, the right side has 2 moles), which also produces more colorless \(\text{N}_2\text{O}_4\) and makes the mixture lighter brown.

Award 1 mark for the correct option A. Reject B, C, and D because they show incorrect applications of Le Chatelier's principle for temperature or pressure changes.

During the electrolysis of concentrated aqueous sodium chloride (brine), chloride ions (\(\text{Cl}^-\)) are discharged at the anode to form chlorine gas (\(\text{Cl}_2\)). Hydrogen ions (\(\text{H}^+\)) are discharged at the cathode in preference to sodium ions, forming hydrogen gas (\(\text{H}_2\)). The remaining sodium ions (\(\text{Na}^+\)) and hydroxide ions (\(\text{OH}^-\)) leave a solution of sodium hydroxide, which is alkaline, so the pH of the electrolyte increases.

Award 1 mark for the correct option A. Reject B, C, and D because they list incorrect electrode products or an incorrect pH trend.

Alkanes are generally unreactive but undergo substitution reactions with halogens in the presence of ultraviolet (UV) light. The UV light provides the activation energy needed to break the covalent bond in chlorine molecules to initiate the reaction, where a chlorine atom replaces a hydrogen atom in methane.

Award 1 mark for the correct option A. Reject B, C, and D as they describe incorrect reaction conditions or reaction types (alkanes do not undergo addition reactions).

First, calculate the number of moles of hydrazine reactants: \(n(\text{N}_2\text{H}_4) = 1.6\text{ g} / 32\text{ g/mol} = 0.05\text{ mol}\). According to the stoichiometry of the reaction, \(1\text{ mol}\) of \(\text{N}_2\text{H}_4\) yields \(1\text{ mol}\) of \(\text{N}_2\) and \(2\text{ mol}\) of \(\text{H}_2\text{O}\). Both products are gases at r.t.p. Total moles of gas produced per mole of hydrazine = \(1 + 2 = 3\text{ moles of gas}\). Total moles of gas produced from \(0.05\text{ mol}\) hydrazine = \(3 \times 0.05\text{ mol} = 0.15\text{ mol}\). Finally, calculate the volume at r.t.p.: \(\text{Volume} = 0.15\text{ mol} \times 24\text{ dm}^3\text{/mol} = 3.6\text{ dm}^3\).

1 mark for the correct option C.

During the electrolysis of concentrated aqueous sodium chloride: At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions because of their high concentration, producing chlorine gas. At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) are discharged in preference to sodium ions (\(\text{Na}^+\)) because hydrogen is lower in the reactivity series, producing hydrogen gas. Since hydrogen ions are removed from the solution around the cathode, an excess of hydroxide ions (\(\text{OH}^-\)) is left behind, making the solution alkaline and causing the pH to increase.

1 mark for the correct option A.

According to Le Chatelier's Principle: 1. Since the forward reaction is exothermic (\(\Delta H < 0\)), increasing the temperature shifts the equilibrium in the endothermic direction (to the left), which decreases the yield of product \(\text{C}\). 2. There are more moles of gas on the reactant side (3 moles) than the product side (2 moles). Decreasing the pressure shifts the equilibrium in the direction of more moles of gas (to the left), which further decreases the yield of product \(\text{C}\). Therefore, high temperature and low pressure yield the lowest amount of \(\text{C}\).

1 mark for the correct option B.

An increase in temperature increases the kinetic energy of the particles. This results in two key effects: The particles move faster, leading to a higher frequency of collisions (Statement 2 is correct). A much larger fraction of the colliding particles possess energy equal to or exceeding the activation energy (Statement 3 is correct). Statement 1 is incorrect because temperature has no effect on the activation energy itself (only a catalyst can lower the activation energy by providing an alternative pathway).

1 mark for the correct option B.

Barium sulfate is an insoluble salt. The standard preparation method for insoluble salts is precipitation. This involves mixing two soluble salts (barium chloride and sodium sulfate), filtering the mixture, washing the residue with distilled water to remove soluble impurities, and drying the purified precipitate.

1 mark for the correct option B.

There are four structural isomers with the molecular formula \(\text{C}_4\text{H}_9\text{Cl}\): 1. 1-chlorobutane: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}\); 2. 2-chlorobutane: \(\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3\); 3. 1-chloro-2-methylpropane: \(\text{(CH}_3\text{)}_2\text{CHCH}_2\text{Cl}\); 4. 2-chloro-2-methylpropane: \(\text{(CH}_3\text{)}_3\text{CCl}\).

1 mark for the correct option C.

During the heating of a pure solid, the first plateau represents melting (solid to liquid transition) and the second plateau represents boiling (liquid to gas transition). During boiling, the temperature stays constant because the thermal energy supplied is entirely used to completely overcome the forces of attraction holding the liquid particles together, rather than increasing the kinetic energy (and thus temperature) of the particles.

1 mark for the correct option B.

Element \(X\) has 12 protons, so its neutral atom has configuration 2, 8, 2. To form a stable ion, it loses its 2 outer-shell electrons to form \(X^{2+}\) with configuration 2, 8. Element \(Y\) has 9 protons, so its neutral atom has configuration 2, 7. To form a stable ion, it gains 1 electron to form \(Y^-\) with configuration 2, 8. Both ions end up with a stable 2, 8 noble gas configuration.

1 mark for the correct option A.

The volume of carbon dioxide produced is equal to the volume of gas absorbed by the sodium hydroxide: 110 cm\(^3\) - 30 cm\(^3\) = 80 cm\(^3\). Since 20 cm\(^3\) of hydrocarbon produced 80 cm\(^3\) of CO\(_2\), each molecule of hydrocarbon contains 4 carbon atoms (x = 4). The volume of unreacted oxygen remaining is 30 cm\(^3\), so the volume of oxygen that reacted is 150 cm\(^3\) - 30 cm\(^3\) = 120 cm\(^3\). The ratio of C\(_4\)H\(_y\) to O\(_2\) reacting is 20 : 120, which is 1 : 6. The balanced equation for the combustion of C\(_4\)H\(_y\) is C\(_4\)H\(_y\) + (4 + y/4)O\(_2\) \(\rightarrow\) 4CO\(_2\) + y/2 H\(_2\)O. Therefore, 4 + y/4 = 6, which gives y = 8. The formula is C\(_4\)H\(_8\).

1 mark for the correct option (D).

In the original experiment, the number of moles of the limiting reactant, HCl, is 50/1000 * 1.0 = 0.050 mol. To obtain the same final volume of carbon dioxide, the number of moles of HCl must remain 0.050 mol. To obtain a steeper initial gradient, the rate of reaction must be faster, which requires a higher concentration of hydrochloric acid. In option C, 25 cm\(^3\) of 2.0 mol/dm\(^3\) HCl has 25/1000 * 2.0 = 0.050 mol of HCl. Because the concentration is higher (2.0 mol/dm\(^3\) vs 1.0 mol/dm\(^3\)), the reaction rate is faster (steeper initial gradient) while producing the same final volume of gas.

1 mark for the correct option (C).

Since the forward reaction is endothermic, the reverse reaction must be exothermic. Cooling the mixture in an ice bath (removing heat) shifts the equilibrium in the exothermic direction (to the left) to oppose the change. This increases the concentration of the pink cobalt complex, turning the solution pink.

1 mark for the correct option (C).

Barium sulfate is an insoluble salt. The standard method of preparing an insoluble salt is precipitation, which involves mixing two soluble salts. Both barium chloride and sodium sulfate are soluble in water. Mixing them produces insoluble barium sulfate and soluble sodium chloride. The precipitate is filtered off, washed with distilled water to remove spectator ions, and dried.

1 mark for the correct option (B).

Isotopes of the same element have the same number of protons and electrons. Since chemical reactions involve the sharing, loss, or gain of electrons, the chemical properties of an element depend solely on its electronic configuration (especially the outer-shell electrons), which is identical for both isotopes.

1 mark for the correct option (C).

The general formula C\(_n\)H\(_{2n}\)O\(_2\) corresponds to both carboxylic acids and esters. Since n = 4 matches this general formula, C\(_4\)H\(_8\)O\(_2\) could represent either a carboxylic acid (such as butanoic acid) or an ester (such as ethyl ethanoate).

1 mark for the correct option (B).

At the positive electrode (anode), chloride ions are in high concentration and are selectively discharged to form chlorine gas. At the negative electrode (cathode), hydrogen ions from water are discharged to form hydrogen gas because hydrogen is less reactive than sodium. This leaves sodium (Na\(^+\)) and hydroxide (OH\(^-\)) ions in the solution, forming sodium hydroxide. Since sodium hydroxide is an alkali, the pH of the remaining solution increases.

1 mark for the correct option (A).

Energy absorbed to break bonds: 1 * (H-H) + 1 * (Cl-Cl) = 436 + 242 = 678 kJ/mol. Energy released when new bonds are formed: 2 * (H-Cl) = 2 * 431 = 862 kJ/mol. Energy change, \(\Delta H\) = Energy absorbed - Energy released = 678 - 862 = -184 kJ/mol.

1 mark for the correct option (A).

1. Calculate the number of moles of lead(II) nitrate: \(\text{moles} = \frac{6.62 \text{ g}}{331 \text{ g/mol}} = 0.02 \text{ mol}\). 2. Determine the molar ratio between lead(II) nitrate and the total gas produced: From the balanced equation, 2 moles of \(\text{Pb(NO}_3)_2\) produce 4 moles of \(\text{NO}_2\) and 1 mole of \(\text{O}_2\), giving a total of 5 moles of gas. The ratio is 2 : 5. 3. Calculate the total moles of gas produced: \(\text{moles of gas} = 0.02 \times \frac{5}{2} = 0.05 \text{ mol}\). 4. Calculate the volume of gas at r.t.p.: \(\text{Volume} = 0.05 \text{ mol} \times 24.0 \text{ dm}^3/\text{mol} = 1.20 \text{ dm}^3\).

1 mark for the correct option C. Award 1 mark for correct calculation of moles of lead(II) nitrate (0.02 mol), identifying the 2:5 ratio of reactant to total gas, and calculating the volume of 1.20 dm³.

Increasing the pressure shifts the equilibrium in the direction of fewer moles of gas. There are 3 moles of gas on the reactant side and 2 moles of gas on the product side, so increasing pressure increases the yield of \(\text{SO}_3\). Decreasing the temperature shifts the equilibrium in the exothermic direction (to the right) to release heat because the forward reaction is exothermic (\(\Delta H\) is negative), so decreasing the temperature also increases the yield of \(\text{SO}_3\).

1 mark for the correct option A. Award 1 mark for identifying that high pressure favours the side with fewer gas moles (RHS) and low temperature favours the exothermic direction (RHS).

At the anode (positive electrode), chloride ions (\(\text{Cl}^-\)) are discharged in preference to hydroxide ions because the solution is concentrated, producing chlorine gas (\(\text{Cl}_2\)). At the cathode (negative electrode), hydrogen ions (\(\text{H}^+\)) are discharged in preference to sodium ions (\(\text{Na}^+\)) because hydrogen is less reactive than sodium, producing hydrogen gas (\(\text{H}_2\)). As \(\text{H}^+\rightleftharpoons \text{H}_2\) proceeds at the cathode, hydroxide ions (\(\text{OH}^-\)) remain in the solution. This increases the concentration of hydroxide ions around the cathode, making the solution alkaline and causing the pH to increase.

1 mark for the correct option B. Award 1 mark for identifying chlorine at the anode, hydrogen at the cathode, and an increase in local pH due to excess hydroxide ions.

The rate of reaction is indicated by the initial gradient of the curve; a steeper curve (Curve Y) indicates a faster reaction rate. The total volume of gas produced is determined by the limiting reactant, which is hydrochloric acid. Because Curve Y levels off at the same volume as Curve X, the amount (moles) of acid must remain unchanged. Using smaller calcium carbonate chips increases the surface area, which increases the rate of reaction (steeper curve) but does not alter the moles of the limiting reactant, so the final volume of gas remains the same. Options B and C would increase the moles of acid, doubling the final volume of gas. Option D would decrease the initial rate of reaction due to the larger chip size.

1 mark for the correct option A. Award 1 mark for explaining that smaller chip size increases rate (steeper gradient) without altering the stoichiometric yield of carbon dioxide.

Copper(II) sulfate is a soluble salt. It cannot be prepared by titration (A) because titration is used to prepare soluble salts from two soluble starting materials (e.g., sodium salts). It cannot be prepared by precipitation (B) because both copper(II) chloride and sulfuric acid are soluble, and copper(II) sulfate is also soluble, so no precipitate forms. Copper metal does not react with dilute sulfuric acid (D) because copper is below hydrogen in the reactivity series. It is prepared by reacting an excess of an insoluble carbonate (copper(II) carbonate) with dilute sulfuric acid (C). Excess solid is filtered off, and the filtrate is crystallised and dried.

1 mark for the correct option C. Award 1 mark for selecting the insoluble carbonate + acid route as the correct preparation method for a soluble copper salt.

Isomers are compounds that share the same molecular formula but have different structural formulae (different structural arrangements of atoms). Because they have different structures, they have different physical properties (such as different boiling points due to differences in surface area and branching) and slightly different chemical properties. Since they have the same molecular formula, they must also have the same empirical formula, making option B the correct statement.

1 mark for the correct option B. Award 1 mark for identifying the correct definition of structural isomers.

The test with aqueous sodium hydroxide produces a green precipitate. Both \(\text{Fe}^{2+}\) and \(\text{Cr}^{3+}\) form green precipitates, but chromium(III) hydroxide is soluble in excess sodium hydroxide to form a green solution, whereas iron(II) hydroxide is insoluble in excess. This confirms the presence of \(\text{Fe}^{2+}\) ions. The test with dilute nitric acid and barium nitrate produces a white precipitate of barium sulfate, confirming the presence of sulfate ions (\(\text{SO}_4^{2-}\)). Therefore, solution X is iron(II) sulfate.

1 mark for the correct option B. Award 1 mark for identifying iron(II) ions from the insoluble green precipitate and sulfate ions from the white precipitate with barium nitrate.

Chlorine is more reactive than bromine because halogen reactivity decreases down Group VII. Therefore, chlorine displaces bromine from potassium bromide to form aqueous bromine and potassium chloride: \(\text{Cl}_2(\text{g}) + 2\text{KBr}(\text{aq}) \rightarrow 2\text{KCl}(\text{aq}) + \text{Br}_2(\text{aq})\). Aqueous bromine is orange-brown, so the colourless solution turns orange-brown. Bromide ions (\(\text{Br}^-\), configuration 2,8,18,8) are oxidised to bromine atoms (configuration 2,8,18,7), meaning statement C is incorrect. Chlorine gas is reduced, making it an oxidising agent rather than a reducing agent, so statement D is incorrect.

1 mark for the correct option A. Award 1 mark for correctly describing the displacement reaction, the colour change, and the roles of the oxidising and reducing agents.

Part (a): Increasing temperature favors the endothermic reaction to absorb the added thermal energy. Since the forward reaction is exothermic, the backward reaction is endothermic. Thus, the equilibrium shifts to the left.

Part (b): Increasing pressure shifts the equilibrium towards the side with fewer moles of gas to reduce the pressure. On the left side there are 3 moles of gas (1 CO + 2 H2), and on the right side there is 1 mole of gas (1 CH3OH). Therefore, the equilibrium shifts to the right, increasing the yield of methanol.

Part (c)(i): A catalyst increases the rate of both the forward and reverse reactions equally.
(ii) It does not change the position of equilibrium because it decreases the activation energy of both forward and reverse reactions by the same amount.

Part (d): The equilibrium expression is \(K_c = \frac{[\text{CH}_3\text{OH(g)}]}{[\text{CO(g)}][\text{H}_2\text{(g)}]^2}\).

Part (e)(i): Moles of \(\text{CO} = \frac{140}{28} = 5.0\text{ mol}\). Molar ratio of \(\text{CO} : \text{CH}_3\text{OH} = 1:1\), so \(5.0\text{ mol}\) of \(\text{CH}_3\text{OH}\) is theoretically produced. Mass of \(\text{CH}_3\text{OH} = 5.0 \times 32 = 160\text{ g}\).
(ii) Percentage yield = \(\frac{112}{160} \times 100\\% = 70.0\\%\).

- Part (a) [3 marks]:
- Equilibrium shifts to the left / reactants side [1]
- Forward reaction is exothermic / backward reaction is endothermic [1]
- System shifts to oppose the temperature rise / absorb heat [1]
- Part (b) [3 marks]:
- Yield of methanol increases / equilibrium shifts to the right [1]
- Fewer moles of gas on the product side / right-hand side [1]
- 3 moles of gas on left vs 1 mole of gas on right [1]
- Part (c) [2 marks]:
- (i) Increases the rates of both reactions equally [1]
- (ii) Lowers activation energy for both reactions by the same amount [1]
- Part (d) [2 marks]:
- Correct expression with products over reactants [1]
- Correct powers/coefficients (squared for H2) [1]
- Part (e) [3.333 marks]:
- (i) Moles of CO = 5.0 mol [1]; Mass of CH3OH = 160 g [1]
- (ii) Yield = 70.0% / 70% [1.333]

Part (a): The flask contains the reacting mixture. A stopper/bung is used to seal the flask and prevent gas escape. A delivery tube directs the gas into a gas syringe or a measuring cylinder inverted over water to measure the volume of gas.

Part (b)(i): Small chips have a larger surface area. This exposes more particles to collisions at any one time, increasing the frequency of collisions and thus the rate of reaction.
(ii) Higher temperature in Experiment 3 increases the rate, meaning the curve is steeper at the start. Since both have the same limiting reactant (5 g of marble chips) and excess acid, the total amount of CO2 produced is the same, so both curves level off at the same volume.

Part (c): 1) Particles gain kinetic energy and move faster, leading to more frequent collisions. 2) More particles have energy greater than or equal to the activation energy, resulting in a higher percentage of successful collisions.

Part (d): Average rate = Volume / Time = \(30\text{ cm}^3 / 20\text{ s} = 1.5\text{ cm}^3/\text{s}\).

- Part (a) [3 marks]:
- Conical flask with reactant mixture and bung/stopper [1]
- Delivery tube connected to gas syringe or inverted measuring cylinder in water trough [1]
- Labels indicating flask, syringe/measuring cylinder, and acid/marble chips [1]
- Part (b) [5 marks]:
- (i) Small chips have larger surface area [1]
- More frequent collisions [1]
- (ii) Experiment 3 has a steeper initial gradient / curve [1]
- Both curves level off at the exact same final volume [1]
- Since the mass of marble chips / limiting reactant is identical, the amount of product is identical [1]
- Part (c) [4 marks]:
- Particles gain kinetic energy and move faster [1]
- More frequent collisions per unit time [1]
- More particles have energy \(\ge\) activation energy [1]
- Higher frequency / proportion of successful collisions [1]
- Part (d) [1.333 marks]:
- Correct calculation: 30 / 20 = 1.5 [1.333]

Part (a): A homologous series is a series of compounds with the same functional group, similar chemical properties, a general formula, and each member differing from the next by a \(-\text{CH}_2-\) unit.

Part (b)(i): Propyl ethanoate is derived from ethanoic acid (supplying the ethanoate part) and propan-1-ol (supplying the propyl part).
(ii): The fully displayed structure must show every bond. Specifically, \(\text{CH}_3-\text{C}(=\text{O})-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3\).
(iii): Catalyst is concentrated sulfuric acid. Reaction requires heating.

Part (c)(i): The mass of the empirical unit \(\text{C}_2\text{H}_4\text{O}\) is \((2 \times 12) + (4 \times 1) + 16 = 44\). Since \(M_r = 88\), the scaling factor is \(88 / 44 = 2\). Therefore, the molecular formula is \(\text{C}_4\text{H}_8\text{O}_2\).
(ii): Possible esters with the formula \(\text{C}_4\text{H}_8\text{O}_2\) are Ethyl ethanoate (\(\text{CH}_3\text{COOCH}_2\text{CH}_3\)), Methyl propanoate (\(\text{CH}_3\text{CH}_2\text{COOCH}_3\)), or Propyl methanoate (\(\text{HCOOCH}_2\text{CH}_2\text{CH}_3\)). Any two of these drawn with their correct IUPAC names are acceptable.

- Part (a) [2 marks]:
- Same functional group / same general formula [1]
- Similar chemical properties / consecutive members differ by a \(-\text{CH}_2-\) unit [1]
- Part (b) [6 marks]:
- (i) Ethanoic acid [1] and propan-1-ol / propanol [1]
- (ii) Correct skeletal arrangement of atoms [1] and all bonds shown explicitly including all C-H, C-O, C=O bonds [1]
- (iii) Concentrated sulfuric acid [1] and heat / reflux [1]
- Part (c) [5.333 marks]:
- (i) Calculation of empirical formula mass = 44 [1], ratio 88/44 = 2 leading to C4H8O2 [1]
- (ii) Drawing of first correct ester structure [1] and its correct IUPAC name [0.5]
- Drawing of second correct ester structure [1] and its correct IUPAC name [0.833]

Part (a):
- Molten lead(II) bromide contains \(\text{Pb}^{2+}\) and \(\text{Br}^-\). At the anode, bromide ions are oxidized to bromine gas. At the cathode, lead ions are reduced to lead metal.
- Concentrated aqueous sodium chloride contains \(\text{Na}^+\), \(\text{H}^+\), \(\text{Cl}^-\), and \(\text{OH}^-\). At the anode, chloride ions are discharged preferentially due to high concentration, forming chlorine gas. At the cathode, hydrogen ions are discharged preferentially over sodium because hydrogen is lower in the reactivity series, forming hydrogen gas.

Part (b)(i): Water/hydroxide ions are oxidized at the anode to form oxygen gas (colorless bubbles). Equation: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\).
(ii): Copper ions are reduced at the cathode to form copper metal (red-brown deposit). Equation: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\).
(iii): The blue color is due to \(\text{Cu}^{2+}\) ions. As they are removed by reduction at the cathode, and not replaced at the anode, their concentration decreases and the blue color fades.

Part (c)(i): With copper electrodes, copper at the anode dissolves (oxidizes to \(\text{Cu}^{2+}\)), so its mass decreases. Copper deposits onto the cathode, so its mass increases.
(ii): The rate of copper dissolution at the anode equals the rate of copper deposition at the cathode. The concentration of \(\text{Cu}^{2+}\) in the electrolyte remains constant, so the intensity of the blue color remains unchanged.

- Part (a) [4 marks, 1 mark each]:
- (i) Bromine / \(\text{Br}_2\) [1]
- (ii) Lead / \(\text{Pb}\) [1]
- (iii) Chlorine / \(\text{Cl}_2\) [1]
- (iv) Hydrogen / \(\text{H}_2\) [1]
- Part (b) [6 marks]:
- (i) Bubbles of gas / effervescence [1] and correct equation: \(4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-\) (or equivalent) [1]
- (ii) Red-brown / pink-brown solid deposited [1] and correct equation: \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\) [1]
- (iii) Blue color fades / becomes lighter [1] because concentration of \(\text{Cu}^{2+}\) decreases as they are discharged [1]
- Part (c) [3.333 marks]:
- (i) Anode mass decreases and cathode mass increases [2]
- (ii) Blue color remains the same [1.333]

Part (a):
1. Add copper(II) oxide to a measured volume of warm dilute sulfuric acid.
2. Ensure copper(II) oxide is in excess (some unreacted solid remains at the bottom) so all acid is neutralized.
3. Filter the mixture to remove excess copper(II) oxide (residue).
4. Heat the filtrate (copper(II) sulfate solution) in an evaporating basin until the crystallization point is reached (crystals start forming on a glass rod).
5. Leave the hot solution to cool slowly so large crystals can form.
6. Filter the crystals, rinse with a tiny volume of cold distilled water, and dry between sheets of filter paper.

Part (b)(i): The preparation of an insoluble salt from two soluble salts is a precipitation reaction.
(ii): Soluble barium salt: barium chloride or barium nitrate. Soluble sulfate: sodium sulfate, potassium sulfate, or sulfuric acid.
(iii): The ionic equation involves only the ions reacting to form the solid precipitate: \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\).

Part (c):
Moles of \(\text{BaCl}_2 = \text{Volume in dm}^3 \times \text{Concentration} = \frac{50.0}{1000} \times 0.200 = 0.0100\text{ mol}\).
Since \(1\text{ mol}\) of \(\text{BaCl}_2\) reacts to produce \(1\text{ mol}\) of \(\text{BaSO}_4\), the moles of \(\text{BaSO}_4\) produced = \(0.0100\text{ mol}\).
Mass of \(\text{BaSO}_4 = 0.0100\text{ mol} \times 233\text{ g/mol} = 2.33\text{ g}\).

- Part (a) [6 marks]:
- Warm the sulfuric acid and add excess copper(II) oxide [1]
- Stir / mix well [1]
- Filter to remove unreacted copper(II) oxide [1]
- Heat the filtrate until crystallization point / saturation point is reached [1]
- Allow to cool to form crystals [1]
- Filter the crystals, wash with a small volume of cold distilled water, and dry with filter paper [1]
- Part (b) [5 marks]:
- (i) Precipitation [1]
- (ii) Barium chloride / barium nitrate [1] and sodium sulfate / potassium sulfate / dilute sulfuric acid [1]
- (iii) \(\text{Ba}^{2+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{BaSO}_4\text{(s)}\) (Correct species [1], correct state symbols [1])
- Part (c) [2.333 marks]:
- Moles of barium chloride = 0.0100 mol [1]
- Mass of barium sulfate = 2.33 g [1.333]

Part (a): Isotopes are atoms of the same element (same atomic number / same number of protons) but with different mass numbers (different numbers of neutrons).

Part (b)(i): Chemical properties are determined by the electronic configuration, specifically the number of outer shell electrons, which is identical in isotopes.
(ii) Chlorine has atomic number 17. So:
- \(\phantom{}^{35}\text{Cl}\) has 17 protons, 17 electrons, and \(35 - 17 = 18\) neutrons.
- \(\phantom{}^{37}\text{Cl}\) has 17 protons, 17 electrons, and \(37 - 17 = 20\) neutrons.
(iii) Let \(x\) be the fractional abundance of \(\phantom{}^{35}\text{Cl}\).
\(35x + 37(1-x) = 35.5\)
\(35x + 37 - 37x = 35.5\)
\(-2x = -1.5 \Rightarrow x = 0.75\).
Thus, the percentage abundance is \(75\\%\).

Part (c)(i): Magnesium (Group II) loses two electrons to form \(\text{Mg}^{2+}\). Chlorine (Group VII) gains one electron to form \(\text{Cl}^-\). The diagram should show one \(\text{Mg}^{2+}\) ion and two \(\text{Cl}^-\) ions. The magnesium ion has no electrons in its outer shell (or shows the full second shell), and each chloride ion has 8 electrons in its outer shell (one electron transferred from magnesium, shown with a different symbol, e.g., dot vs cross) and a single negative charge.
(ii) Magnesium chloride has a giant ionic lattice. There are strong electrostatic attractions between the oppositely charged \(\text{Mg}^{2+}\) and \(\text{Cl}^-\) ions, which require a large amount of thermal energy to overcome, resulting in a high melting point.

- Part (a) [2 marks]:
- Atoms of same element / same proton (or atomic) number [1]
- Different numbers of neutrons (or mass number) [1]
- Part (b) [7 marks]:
- (i) Same number of outer shell electrons [1]
- (ii) \(\phantom{}^{35}\text{Cl}\): 17 p, 18 n, 17 e [1.5]
- \(\phantom{}^{37}\text{Cl}\): 17 p, 20 n, 17 e [1.5] (All 6 correct = 3 marks, 4-5 correct = 2 marks, 2-3 correct = 1 mark)
- (iii) Setting up equation: \(35(x) + 37(100-x) = 35.5 \times 100\) [1]
- Correct rearrangement: \(2x = 150\) [1]
- Correct answer: 75% [1]
- Part (c) [4.333 marks]:
- (i) Magnesium ion shown as \(\text{Mg}^{2+}\) [1]
- Two chloride ions shown as \(\text{Cl}^-\), each with 8 outer electrons (crosses/dots distinguished) [1]
- Ratio of ions is 1:2 and correct charges on both [1]
- (ii) Giant ionic structure with strong electrostatic attraction between oppositely charged ions [1.333]

An exemplar plan:
1. **Preparation**: Take four identical iron nails. Keep one nail uncoated as a control. Use a paintbrush to apply coating A to the second nail, coating B to the third nail, and coating C to the fourth nail. Ensure the entire surface of each nail is completely and evenly coated. Allow them to dry.
2. **Setup**: Place each nail into a separate, labeled test-tube. Add an equal volume of distilled water (e.g., 5 cm³) to each test-tube so that the nails are partially submerged, exposing them to both water and air.
3. **Variables to control**: Keep the volume of water the same, the type and size of the test-tubes the same, the size and material of the nails the same, and store all test-tubes in the same environment (same temperature, light exposure).
4. **Observations**: Leave the test-tubes for two weeks. Inspect the nails daily or at the end of the two weeks. Record the appearance of rust (red-brown solid) on each nail.
5. **Evaluation**: Compare the amount or area of rust on each coated nail. Alternatively, weigh each nail before coating (without coating) and after removing rust at the end, though visual comparison of the surface area rusted or time taken for rust to first appear is highly practical. The coating on the nail with the least amount of rust (or no rust) is the most effective.

Award marks as follows (up to 10 marks):
- **Method (Max 5 marks)**:
- [1] Coating the nails: Describes coating separate identical nails with A, B, and C using a paintbrush and letting them dry.
- [1] Control setup: Sets up an uncoated control nail.
- [1] Exposing to rust conditions: Places each nail in a test-tube with water (distilled water) and air.
- [1] Duration: Leave the setups for a specified suitable time frame (e.g., several days / 1 to 2 weeks).
- [1] Detail: Ensure the nails are fully or partially submerged under identical conditions.
- **Control Variables (Max 2 marks)**:
- [1] Mentioning at least one controlled physical variable (e.g., identical nails, same volume of water).
- [1] Mentioning environmental control (e.g., same temperature, same exposure to air/location).
- **Observations / Measurements (Max 2 marks)**:
- [1] Visual observation: Observe and record the amount/area of red-brown rust on each nail (or rate/time taken for rust to appear).
- [1] Alternative quantitative method: Weighing dry nails before and after rusting (after removing rust) to find mass loss/gain, or estimating percentage surface area covered by rust.
- **Conclusion (Max 1 mark)**:
- [1] The most effective coating is the one that shows the least amount of rust (or zero rust) at the end of the period.

(a) (i) The volume halfway between \(24.0\text{ cm}^3\) and \(26.0\text{ cm}^3\) is \(25.0\text{ cm}^3\).
(ii) At \(90\text{ s}\), the volume is exactly \(54.0\text{ cm}^3\).

(b) Variables to keep constant: The concentration of hydrochloric acid or the volume of hydrochloric acid (or the mass of calcium carbonate). Explaining: Changing acid concentration changes the rate of reaction, which would interfere with the variable being tested (lump size).

(c) The reaction is finished when the volume of gas stops increasing / remains constant over consecutive readings. This is because one of the reactants (the limiting reactant, which is calcium carbonate since acid is in excess) has been completely used up, so no more gas particles can be produced.

(d) At the start, the concentration of reactant particles (specifically \(\text{H}^+\) ions in the acid) is at its highest. Therefore, there is the maximum frequency of successful collisions between reactant particles per unit time.

(e) To investigate acid concentration: Keep the mass and size of calcium carbonate lumps constant. Repeat the experiment using a different concentration of hydrochloric acid (e.g., \(1.0\text{ mol/dm}^3\) instead of \(2.0\text{ mol/dm}^3\)), keeping the temperature and volume of acid constant. Expected observation: With a higher concentration, the initial slope of the volume-time graph will be steeper, indicating a faster rate of reaction.

Award marks as follows:
- **(a) Readings (2 marks)**:
- [1] (i) \(25.0\text{ cm}^3\) (accept \(25\))
- [1] (ii) \(54.0\text{ cm}^3\) (accept \(54\))
- **(b) Controlled variable (2 marks)**:
- [1] State: Concentration of hydrochloric acid / volume of acid / mass of calcium carbonate.
- [1] Explain: If these are changed, they also affect the rate of reaction / total volume of gas, making it an unfair test.
- **(c) Completion (2 marks)**:
- [1] Observation: The volume of gas stops increasing / stays constant (horizontal line on a graph).
- [1] Explanation: The limiting reactant (calcium carbonate) is completely used up.
- **(d) Initial rate (2 marks)**:
- [1] High concentration: Reactant concentration/particles are at their maximum at the start.
- [1] Collisions: Highest frequency of collisions per unit time (or most frequent successful collisions).
- **(e) Modification & Result (2 marks)**:
- [1] Modification: Use different concentrations of hydrochloric acid, keeping mass of \(\text{CaCO}_3\) and volume of acid constant.
- [1] Observation: A higher concentration results in a faster rate (steeper initial curve/shorter time to complete).

(a) The gas that turns limewater cloudy is carbon dioxide (\(\text{CO}_2\)).

(b) The reaction is thermal decomposition (the breakdown of a substance by heat).

(c) The black powder is copper(II) oxide (\(\text{CuO}\)). When nitric acid is added to it, it forms a blue solution (Solution Y), which is typical of copper(II) ions in aqueous solution (\(\text{Cu}^{2+}\text{(aq)}\)).

(d) Since carbon dioxide gas was released upon heating, the anion present in Salt X is the carbonate ion, \(\text{CO}_3^{2-}\).

(e) (i) With aqueous sodium hydroxide: A light blue precipitate of copper(II) hydroxide is formed, which is insoluble in excess.
(ii) With aqueous ammonia: A light blue precipitate is formed, which dissolves in excess ammonia to give a deep blue solution.

(f) To test for the presence of water: Add anhydrous copper(II) sulfate, which turns from white to blue (or add anhydrous cobalt(II) chloride paper, which turns from blue to pink).

Award marks as follows:
- **(a) Gas identification (1 mark)**:
- [1] Carbon dioxide (accept \(\text{CO}_2\)).
- **(b) Reaction type (1 mark)**:
- [1] Thermal decomposition.
- **(c) Black powder identity & explanation (2 marks)**:
- [1] Copper(II) oxide / \(\text{CuO}\).
- [1] Explanation: Copper(II) compounds are typically blue/green in solution, and copper(II) oxide is black and reacts with acid to form a blue \(\text{Cu}^{2+}\) solution.
- **(d) Anion formula (1 mark)**:
- [1] \(\text{CO}_3^{2-}\) (must have correct charge and formula; reject 'carbonate' as chemical formula is asked).
- **(e) Qualitative tests (4 marks)**:
- [1] (i) Dropwise/Excess NaOH: Light blue precipitate.
- [1] insoluble in excess NaOH.
- [1] (ii) Dropwise/Excess Ammonia: Light blue precipitate.
- [1] dissolves in excess to form a deep blue solution.
- **(f) Water test (1 mark)**:
- [1] Anhydrous copper(II) sulfate turns white to blue (or anhydrous cobalt(II) chloride turns blue to pink).

(a) The acid is warmed to increase the rate of the reaction (speed it up).

(b) The student will know that copper(II) oxide is in excess because:
1. A black solid/powder remains visible at the bottom of the beaker and does not dissolve even after thorough stirring.
2. The solution stops getting deeper blue / the color remains constant.

(c) (i) Residue: Unreacted copper(II) oxide (\(\text{CuO}\))
(ii) Filtrate: Copper(II) sulfate solution (\(\text{CuSO}_4\text{(aq)}\))

(d) Step 5 is incorrect because heating the solution to complete dryness will remove the water of crystallisation, resulting in a powdery white mass of anhydrous copper(II) sulfate rather than blue hydrated crystals.
To obtain large crystals:
1. Heat the filtrate in an evaporating dish to evaporate some water until the crystallization point is reached (or when crystals start to form on a cold glass rod).
2. Leave the hot, saturated solution to cool slowly at room temperature (which allows large crystals to grow).
3. Filter off the crystals from the remaining mother liquor.

(e) Dry the crystals by gently patting/pressing them between sheets of filter paper (or leaving them in a warm oven/desiccator, but not direct high heat which would dehydrate them).

Award marks as follows:
- **(a) Warming (1 mark)**:
- [1] To increase the rate of reaction / make it react faster.
- **(b) Observations for excess (2 marks)**:
- [1] Black solid/powder stops dissolving / remains at the bottom of the beaker.
- [1] Blue color of the solution stops deepening / remains constant.
- **(c) Residue and Filtrate (2 marks)**:
- [1] Residue: Copper(II) oxide / \(\text{CuO}\) (accept 'unreacted black powder').
- [1] Filtrate: Copper(II) sulfate solution / \(\text{CuSO}_4\text{(aq)}\).
- **(d) Crystallisation procedure (4 marks)**:
- [1] Critique: Heating to dryness removes water of crystallisation / forms anhydrous powder (or small, poorly formed crystals).
- [1] Saturated solution: Heat/evaporate until the crystallization point is reached / some water is evaporated.
- [1] Slow cooling: Leave to cool slowly to allow crystal growth.
- [1] Isolation: Filter off the crystals from the remaining solution.
- **(e) Drying (1 mark)**:
- [1] Pat dry with filter paper / place in a desiccator / warm oven (reject 'heat with Bunsen burner').