Edexcel IGCSE · Thinka-original Practice Paper

2024 Edexcel IGCSE Chemistry Practice Paper with Answers

Thinka Jun 2024 Cambridge International A Level-Style Mock — Chemistry

180 marks195 mins2024

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 Cambridge International A Level Chemistry paper. Not affiliated with or reproduced from Cambridge.

Paper 1C

Answer all questions. Show all steps in calculations and state units. Calculators and rulers are permitted.

12 Question · 135 marks

Question 1 · Recall and matching tables

7 marks

A student is carrying out chemical tests to identify various cations and anions.

Complete the table by identifying the missing details represented by letters **A** to **G**.

| Ion | Test Reagent / Method | Observation |
| :--- | :--- | :--- |
| Lithium, $ \text{Li}^+ $ | Flame test | **A** |
| Copper(II), $ \text{Cu}^{2+} $ | Flame test | **B** |
| **C** | Add aqueous sodium hydroxide | Green precipitate |
| Iron(III), $ \text{Fe}^{3+} $ | Add aqueous sodium hydroxide | **D** |
| Carbonate, $ \text{CO}_3^{2-} $ | **E** | Effervescence (gas produced turns limewater cloudy) |
| **F** | Add dilute nitric acid and aqueous silver nitrate | Cream precipitate |
| Sulfate, $ \text{SO}_4^{2-} $ | Add dilute hydrochloric acid and aqueous **G** | White precipitate |

Show answer & marking scheme

Worked solution

To complete the table, we recall the standard qualitative tests for ions in Pearson Edexcel IGCSE Chemistry:
- **A**: Lithium ions ($ \text{Li}^+ $) impart a red or crimson colour to a flame.
- **B**: Copper(II) ions ($ \text{Cu}^{2+} $) impart a blue-green (or green-blue) colour to a flame.
- **C**: Aqueous sodium hydroxide reacts with iron(II) ions ($ \text{Fe}^{2+} $) to form a green precipitate of iron(II) hydroxide.
- **D**: Aqueous sodium hydroxide reacts with iron(III) ions ($ \text{Fe}^{3+} $) to form a brown (or red-brown) precipitate of iron(III) hydroxide.
- **E**: Carbonate ions ($ \text{CO}_3^{2-} $) are detected by adding a dilute acid (such as dilute hydrochloric acid), which causes effervescence of carbon dioxide gas.
- **F**: The halide ion that produces a cream precipitate of silver bromide with acidified silver nitrate is bromide ($ \text{Br}^- $).
- **G**: Sulfate ions ($ \text{SO}_4^{2-} $) are tested by adding dilute hydrochloric acid followed by barium chloride solution (or dilute nitric acid followed by barium nitrate solution) to yield a white precipitate of barium sulfate.

Marking scheme

- **A**: 1 mark for red / crimson (reject: orange / pink).
- **B**: 1 mark for blue-green / green-blue (accept: green / blue).
- **C**: 1 mark for iron(II) / $ \text{Fe}^{2+} $ (reject: iron / iron(III) / $ \text{Fe}^{3+} $).
- **D**: 1 mark for brown precipitate / red-brown precipitate / brown solid (reject: rust alone, red alone, or yellow).
- **E**: 1 mark for adding (dilute) acid / (dilute) hydrochloric acid / (dilute) nitric acid (accept: formula $ \text{HCl} $ / $ \text{HNO}_3 $; reject: sulfuric acid).
- **F**: 1 mark for bromide / $ \text{Br}^- $ (reject: bromine / $ \text{Br}_2 $).
- **G**: 1 mark for barium chloride (solution) / $ \text{BaCl}_2 $ (accept: barium nitrate / $ \text{Ba(NO}_3)_2 $).

Question 2 · structured

9 marks

A student investigates the reactivity of metals using various reactions. (a) The student adds a piece of magnesium ribbon to an excess of dilute sulfuric acid. (i) State two observations the student would make during this reaction. (2) (ii) Write a chemical equation, including state symbols, for this reaction. (2) (b) In a second experiment, the student places a piece of zinc into a solution of copper(II) sulfate. A displacement reaction occurs. (i) Explain, in terms of electron transfer, why this reaction is classified as a redox reaction. (3) (ii) State the color change of the solution during this displacement reaction. (2)

Show answer & marking scheme

Worked solution

(a)(i) Any two of the following observations: 1. Bubbles of gas / effervescence / fizzing. 2. The magnesium ribbon dissolves / disappears / gets smaller. 3. The tube becomes warm / temperature increases. (a)(ii) $ \text{Mg(s)} + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{MgSO}_4(\text{aq}) + \text{H}_2(\text{g}) $ (b)(i) Zinc atoms lose electrons to form zinc ions ($ \text{Zn}^{2+} $), which is oxidation. Copper ions ($ \text{Cu}^{2+} $) gain electrons to form copper atoms ($ \text{Cu} $), which is reduction. Since both oxidation and reduction occur simultaneously, it is a redox reaction. (b)(ii) The solution changes from blue to colorless.

Marking scheme

Part (a)(i) [Max 2 marks]:
- M1: Effervescence / bubbles / fizzing (1)
- M2: Magnesium ribbon dissolves / disappears / gets smaller OR the mixture gets warm / temperature increases (1)
- ACCEPT: magnesium reacts to form a colorless solution
- REJECT: 'gas is produced' without mentioning bubbles/fizzing

Part (a)(ii) [Max 2 marks]:
- M1: Correct chemical formulae for reactants and products: $ \text{Mg} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2 $ (1)
- M2: Correct state symbols: $ \text{(s)} $, $ \text{(aq)} $, $ \text{(aq)} $, $ \text{(g)} $ (1) [This mark is dependent on M1 or a very minor formula error]

Part (b)(i) [Max 3 marks]:
- M1: Zinc / $ \text{Zn} $ loses electrons (and is oxidised) (1)
- M2: Copper ions / $ \text{Cu}^{2+} $ gain electrons (and are reduced) (1)
- M3: (Redox because) oxidation and reduction occur together / at the same time (1)
- Note: award 1 mark max out of M1 and M2 if student says 'copper gains electrons' instead of 'copper ions gain electrons'.

Part (b)(ii) [Max 2 marks]:
- M1: Blue (1)
- M2: to colorless (1)
- REJECT: 'clear' instead of 'colorless' for M2

Question 3 · structural

10 marks

Gallium exists as two stable isotopes, $^{69}\text{Ga}$ and $^{71}\text{Ga}$.

(a) State the meaning of the term *isotopes*. (2)

(b) The atomic number of gallium is 31.
(i) Copy and complete the table to show the number of protons, neutrons, and electrons in a neutral atom of each isotope. (3)

| Isotope | Number of protons | Number of neutrons | Number of electrons |
| :--- | :--- | :--- | :--- |
| $^{69}\text{Ga}$ | | | |
| $^{71}\text{Ga}$ | | | |

(ii) A sample of gallium has the following percentage composition by mass:
- $^{69}\text{Ga} = 60.11\%$
- $^{71}\text{Ga} = 39.89\%$

Calculate the relative atomic mass ($A_{\text{r}}$) of this sample of gallium. Give your answer to two decimal places. (3)

(c) Explain, in terms of their electronic configuration, why both isotopes of gallium have identical chemical properties. (2)

Show answer & marking scheme

Worked solution

(a) Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons.

(b) (i)
- The atomic number of Gallium is 31, which means any gallium atom has 31 protons.
- In a neutral atom, the number of electrons equals the number of protons, so both isotopes have 31 electrons.
- The number of neutrons is found by subtracting the atomic number (31) from the mass number:
- For $^{69}\text{Ga}$: $69 - 31 = 38$ neutrons.
- For $^{71}\text{Ga}$: $71 - 31 = 40$ neutrons.

Completed table:
| Isotope | Number of protons | Number of neutrons | Number of electrons |
| :--- | :--- | :--- | :--- |
| $^{69}\text{Ga}$ | 31 | 38 | 31 |
| $^{71}\text{Ga}$ | 31 | 40 | 31 |

(b) (ii)
\[A_{\text{r}} = \frac{(69 \times 60.11) + (71 \times 39.89)}{100}\]
\[A_{\text{r}} = \frac{4147.59 + 2832.19}{100}\]
\[A_{\text{r}} = \frac{6979.78}{100} = 69.7978\]
Rounding to 2 decimal places gives:
\[A_{\text{r}} = 69.80\]

(c) Chemical reactions are determined by the arrangement of electrons, specifically the number of outer shell (valence) electrons. Since both isotopes of gallium have the same atomic number (31), they have identical electronic configurations (and thus both have 3 outer-shell electrons). Consequently, they react in the exact same way.

Marking scheme

(a)
- M1: Atoms (of the same element) with the same number of protons / same atomic number (1)
- M2: but different number of neutrons / different mass number (1)
[Reject: 'different relative atomic masses' for M2]

(b)(i)
- M1: 31 protons and 31 electrons for both isotopes (1)
- M2: 38 neutrons for $^{69}\text{Ga}$ (1)
- M3: 40 neutrons for $^{71}\text{Ga}$ (1)

(b)(ii)
- M1: Expression showing multiplication of mass numbers by their relative percentages and adding them together: $(69 \times 60.11) + (71 \times 39.89)$ or $4147.59 + 2832.19$ (1)
- M2: Division of the sum by 100 (1)
- M3: Correct final value calculated to 2 decimal places: 69.80 (1)
[Award 3 marks for correct final answer with no working shown]
[If 69.7978 or 69.8 is given without correct rounding to 2 d.p., award maximum of 2 marks]

(c)
- M1: (Both isotopes have the) same electronic configuration / same number of outer-shell electrons (1)
- M2: Chemical reactions / properties depend on outer-shell electrons (1)

Question 4 · structural

10 marks

Show answer & marking scheme

Worked solution

Marking scheme

Question 5 · structured

12 marks

An organic compound, X, contains 24.2% carbon, 4.0% hydrogen and 71.8% chlorine by mass.

(a) (i) Calculate the empirical formula of compound X. (3)
(ii) The relative formula mass ($M_r$) of compound X is 99.0. Determine the molecular formula of compound X. (2)

(b) Compound X has a simple molecular structure with a boiling point of 83 °C. Sodium chloride ($NaCl$) is an ionic compound with a boiling point of 1413 °C.
Explain, in terms of structure and bonding, why compound X has a much lower boiling point than sodium chloride. (4)

(c) Draw a dot-and-cross diagram to show the covalent bonding in a molecule of water ($H_2O$). Show outer shell electrons only. (3)

Show answer & marking scheme

Worked solution

(a) (i)
- Moles of C = $24.2 / 12 = 2.017$
- Moles of H = $4.0 / 1 = 4.000$
- Moles of Cl = $71.8 / 35.5 = 2.023$
- Dividing by the smallest value (2.017):
- C = $2.017 / 2.017 = 1$
- H = $4.000 / 2.017 = 1.98 \approx 2$
- Cl = $2.023 / 2.017 = 1.003 \approx 1$
- Empirical formula is $CH_2Cl$.

(ii)
- Empirical formula mass of $CH_2Cl = 12 + (2 \times 1) + 35.5 = 49.5$
- Scale factor = $99.0 / 49.5 = 2$
- Molecular formula = $2 \times CH_2Cl = C_2H_4Cl_2$.

(b)
- Compound X has a simple molecular structure with weak intermolecular forces (forces of attraction between molecules).
- These weak forces require very little thermal energy to overcome, resulting in a low boiling point.
- Sodium chloride has a giant ionic lattice structure with strong electrostatic forces of attraction between oppositely charged ions ($Na^+$ and $Cl^-$).
- These strong ionic bonds require a large amount of energy to break, resulting in a high boiling point.

(c)
- A water molecule has one oxygen atom in the center forming single covalent bonds with two hydrogen atoms.
- The diagram must show two shared pairs of electrons (each containing one dot and one cross) representing the two O-H covalent bonds.
- The oxygen atom must have 4 remaining non-bonding electrons (2 lone pairs) in its outer shell to complete its octet.
- Each hydrogen atom must have only the 2 shared electrons in its outer shell.

Marking scheme

(a) (i)
- M1: Dividing percentages by correct relative atomic masses ($24.2/12$, $4.0/1$, $71.8/35.5$) [1]
- M2: Finding simplest whole-number ratio of 1 : 2 : 1 [1]
- M3: Deducing correct empirical formula of $CH_2Cl$ [1]

(ii)
- M1: Calculating empirical formula mass of 49.5 [1]
- M2: Deducing correct molecular formula $C_2H_4Cl_2$ [1]

(b)
- M1: Identifies compound X has weak intermolecular forces / weak forces between molecules [1]
- M2: Mentions that little energy is needed to break / overcome these weak intermolecular forces [1]
- M3: Identifies sodium chloride has a giant ionic lattice with strong electrostatic forces of attraction between oppositely charged ions (or strong ionic bonds) [1]
- M4: Mentions that a large amount of energy is required to break / overcome these strong bonds [1]

(c)
- M1: Two shared pairs of electrons shown between oxygen and each of the two hydrogen atoms [1]
- M2: Four non-bonding outer-shell electrons shown on the oxygen atom [1]
- M3: No other outer-shell electrons shown on the hydrogen atoms [1]

Question 6 · multi-part

9 marks

A student investigates the food dyes present in a green candy shell using paper chromatography.

They draw a baseline on a piece of chromatography paper, place a spot of the green food dye on the line, and suspend the paper in a beaker containing a solvent.

**(a)** Explain why the student should use a pencil to draw the baseline rather than an ink pen. (2)

**(b)** Explain why the level of the solvent in the beaker must be below the baseline. (2)

**(c)** In the completed chromatogram, the solvent front travels $8.0\text{ cm}$ from the baseline. A blue spot from the green food dye travels $5.2\text{ cm}$. Calculate the $R_{\text{f}}$ value of the blue dye, giving your answer to 2 decimal places. Show your working. (3)

**(d)** The green food dye separates into only a blue spot and a yellow spot. Explain what this reveals about the composition of the green food dye. (2)

Show answer & marking scheme

Worked solution

**(a)** Pencil lead (graphite) is insoluble in the solvent, so it will not dissolve or travel up the chromatography paper. Ink from a pen is soluble in the solvent, so it would run, separate, and interfere with the chromatogram.

**(b)** If the solvent level is above the baseline, the spot of green dye will dissolve directly into the solvent in the beaker rather than traveling up the paper.

**(c)** Use the formula:
\[R_{\text{f}} = \frac{\text{distance travelled by the substance}}{\text{distance travelled by the solvent front}}\]
\[R_{\text{f}} = \frac{5.2\text{ cm}}{8.0\text{ cm}} = 0.65\]

**(d)** The green food dye is a mixture (not a pure substance) because it separates into more than one spot (specifically blue and yellow dyes).

Marking scheme

**Part (a):**
* **M1:** Pencil is insoluble / graphite does not dissolve in the solvent (1)
* **M2:** Pen ink is soluble / would dissolve/run / would interfere with the results (1)

**Part (b):**
* **M1:** To prevent the dye spot on the baseline from dissolving into the solvent / washing off into the beaker (1)
* **M2:** (If it dissolves in the beaker) it will not travel up the paper / chromatogram will not form (1)

**Part (c):**
* **M1:** State formula or show substitution: $\frac{5.2}{8.0}$ (1)
* **M2:** Evaluate fraction: $0.65$ (1)
* **M3:** Correct answer to 2 decimal places: $0.65$ (1)
*(Note: An answer of 0.65 with no working scores 3 marks. 0.7 or other rounding errors with correct working scores 2 marks.)*

**Part (d):**
* **M1:** The green food dye is a mixture (1)
* **M2:** Because it separates into more than one spot / two different dyes / blue and yellow dyes (1)

Question 7 · structured

13 marks

This question is about Group 2 elements and the determination of the formula of a hydrated salt.

(a) Magnesium is a Group 2 metal.

(i) State the trend in reactivity of the Group 2 metals with dilute hydrochloric acid from magnesium to barium. [1]

(ii) Explain this trend in terms of electronic configurations and atomic structure. [3]

(b) A student carries out an experiment to find the value of $x$ in hydrated magnesium sulfate, $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$.

The student heats a sample of the hydrated salt in a crucible using the following method:
- Weigh an empty crucible.
- Add a sample of hydrated magnesium sulfate and weigh again.
- Heat the crucible and its contents strongly using a Bunsen burner.
- Allow the crucible to cool and weigh it.
- Repeat the heating, cooling, and weighing process.

(i) Explain why the student repeats the heating, cooling, and weighing process until a constant mass is obtained. [2]

(ii) The student obtains the following results:
- Mass of empty crucible = $22.54\text{ g}$
- Mass of crucible + $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$ = $27.46\text{ g}$
- Mass of crucible + anhydrous $\text{MgSO}_4$ after heating to constant mass = $24.94\text{ g}$

Use these results to calculate the value of $x$ in the formula $\text{MgSO}_4 \cdot x\text{H}_2\text{O}$.
Show all steps in your calculation.
$[A_r\text{ of Mg} = 24, A_r\text{ of S} = 32, A_r\text{ of O} = 16, A_r\text{ of H} = 1]$ [7]

Show answer & marking scheme

Worked solution

(a)(i) Reactivity increases down the group from magnesium to barium.

(a)(ii) Down the group, the atomic radius increases as there are more occupied electron shells. This means the outer shell electrons are further from the nucleus and experience more shielding from inner electron shells. Consequently, there is a weaker electrostatic attraction between the positive nucleus and the outer electrons, making it easier to lose the two outer electrons to form $2+$ ions.

(b)(i) Heating to constant mass ensures that all the water of crystallisation has been completely driven off (evaporated) from the hydrated salt, ensuring the reaction/decomposition is complete.

(b)(ii)
1. Calculate mass of anhydrous $\text{MgSO}_4$:
$24.94\text{ g} - 22.54\text{ g} = 2.40\text{ g}$

2. Calculate mass of water lost:
$27.46\text{ g} - 24.94\text{ g} = 2.52\text{ g}$

3. Calculate molar masses:
- $M_r(\text{MgSO}_4) = 24 + 32 + (16 \times 4) = 120$
- $M_r(\text{H}_2\text{O}) = (1 \times 2) + 16 = 18$

4. Calculate moles of each species:
- $\text{Moles of } \text{MgSO}_4 = \frac{2.40}{120} = 0.020\text{ mol}$
- $\text{Moles of } \text{H}_2\text{O} = \frac{2.52}{18} = 0.140\text{ mol}$

5. Determine the simplest molar ratio:
$\text{Ratio of } \text{H}_2\text{O} : \text{MgSO}_4 = \frac{0.140}{0.020} = 7$

6. State the value of $x$:
$x = 7$

Marking scheme

(a)(i)
- Reactivity increases / reaction becomes more vigorous (1)

(a)(ii)
- Outer shell electrons are further from the nucleus / more electron shells / atomic radius increases (1)
- More shielding (from inner shells) (1)
- Weaker attraction between the positive nucleus and outer electrons, so outer electrons are lost more easily (1)

(b)(i)
- To ensure all the water of crystallisation has been removed / driven off (1)
- So that the thermal decomposition is complete / the mass of anhydrous salt remains constant (1)

(b)(ii)
- Calculate mass of anhydrous salt = $2.40\text{ g}$ (1)
- Calculate mass of water = $2.52\text{ g}$ (1)
- Calculate molar mass of $\text{MgSO}_4$ as 120 and $\text{H}_2\text{O}$ as 18 (1)
- Calculate moles of $\text{MgSO}_4$ = $0.02\text{ mol}$ (1)
- Calculate moles of $\text{H}_2\text{O}$ = $0.14\text{ mol}$ (1)
- Divide both moles by the smallest value ($0.02$) to find the mole ratio of $1 : 7$ (1)
- State the final integer value of $x = 7$ (1)

[Award full marks for a correct final answer of 7 with working shown]

Question 8 · structured

13 marks

This question is about gases in the atmosphere, covalent bonding, and chemical tests.

(a) Air is a mixture of gases.
(i) State the percentage by volume of nitrogen and oxygen in dry, unpolluted air.
Percentage of nitrogen: ............
Percentage of oxygen: ............
(2)
(ii) Name another gas present in unpolluted air that has a percentage by volume of approximately 0.93%.
(1)

(b) Carbon dioxide is a covalent compound found in the atmosphere.
(i) Draw a dot-and-cross diagram to show the bonding in a molecule of carbon dioxide, $ \text{CO}_2 $. Show only the outer shell electrons.
(3)
(ii) Carbon dioxide can be prepared in the laboratory by reacting dilute hydrochloric acid with calcium carbonate. Write a chemical equation for this reaction. Include state symbols.
(2)

(c) Describe a chemical test to show that a gas is carbon dioxide. Include the observation for a positive result.
(2)

(d) Describe a chemical test to show that a gas is oxygen. Include the observation for a positive result.
(2)

(e) State one environmental problem associated with the increasing percentage of carbon dioxide in the atmosphere.
(1)

Show answer & marking scheme

Worked solution

(a)(i) Nitrogen makes up approximately 78% of dry, unpolluted air, while oxygen makes up approximately 21%.
(ii) Argon is the noble gas present at approximately 0.93%.

(b)(i) In $ \text{CO}_2 $, the carbon atom shares two pairs of electrons with each of the two oxygen atoms to form two double covalent bonds. Carbon (Group 4) has no lone pairs left in its outer shell. Each oxygen (Group 6) has two lone pairs (4 non-bonding electrons) remaining.
(ii) The reaction of calcium carbonate and hydrochloric acid produces calcium chloride, water, and carbon dioxide:
$ \text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} $

(c) Carbon dioxide is tested by bubbling the gas through limewater (aqueous calcium hydroxide). A positive result is indicated when the limewater turns cloudy or milky.

(d) Oxygen is tested by placing a glowing splint into a tube of the gas. The splint will relight in the presence of oxygen.

(e) Increased levels of carbon dioxide contribute to the enhanced greenhouse effect, leading to global warming, climate change, or ocean acidification.

Marking scheme

(a)(i)
- 1 mark for nitrogen: 78% (accept 78% to 79%)
- 1 mark for oxygen: 21% (accept 20.5% to 21%)

(a)(ii)
- 1 mark for Argon / Ar (reject other noble gases)

(b)(i)
- M1: Two pairs of shared electrons (4 electrons) shown between the central C atom and each O atom (double bonds) (1)
- M2: Carbon atom showing 8 outer electrons in total and no lone pairs (1)
- M3: Both oxygen atoms showing 8 outer electrons in total (including 4 non-bonding electrons / 2 lone pairs on each oxygen) (1)

(b)(ii)
- M1: Correct formulae for all reactants and products: $ \text{CaCO}_3 $, $ \text{HCl} $, $ \text{CaCl}_2 $, $ \text{H}_2\text{O} $, $ \text{CO}_2 $ (1)
- M2: Fully balanced equation with correct state symbols: $ \text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} $ (1)

(c)
- Test: bubble the gas through limewater / calcium hydroxide solution (1)
- Observation: turns cloudy / milky / white precipitate forms (1)

(d)
- Test: insert a glowing splint (1)
- Observation: splint relights / reignites (1)

(e)
- 1 mark for any one from: greenhouse effect / global warming / climate change / rising sea levels / ocean acidification

Question 9 · written

15 marks

This question is about hydrocarbons and polymers.

An unsaturated hydrocarbon, A, has the molecular formula $\text{C}_4\text{H}_8$.

(a) (i) State the names of two isomers of $\text{C}_4\text{H}_8$ that are alkenes. (2)
(ii) Draw the displayed formula of each of these isomers. (2)

(b) (i) But-1-ene can undergo polymerization to form poly(but-1-ene). Draw the structure of the repeat unit of poly(but-1-ene). (2)
(ii) State one environmental problem associated with disposing of addition polymers in landfill sites, and one environmental problem caused by their disposal by combustion. (2)

(c) But-1-ene burns completely in oxygen.
(i) Write a balanced chemical equation for the complete combustion of but-1-ene, $\text{C}_4\text{H}_8$. (2)
(ii) A sample containing 1.40 g of but-1-ene is completely burned in excess oxygen. Calculate the volume, in $\text{cm}^3$, of carbon dioxide gas produced at room temperature and pressure (rtp).
[Molar volume of a gas at rtp = 24.0 $\text{dm}^3/\text{mol}$; $M_r$ of $\text{C}_4\text{H}_8 = 56.0$] (3)

(d) Describe a chemical test to distinguish butane from but-1-ene. Include the observations for each compound. (2)

Show answer & marking scheme

Worked solution

(a) (i) The two alkene isomers are but-1-ene and but-2-ene (or 2-methylpropene).
(ii) The displayed formulae must show every atom and every single and double bond:
- But-1-ene: $\text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_3$ drawn with all C-H, C=C, and C-C bonds visible.
- But-2-ene: $\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3$ drawn with all C-H, C=C, and C-C bonds visible.

(b) (i) The repeat unit of poly(but-1-ene) consists of a two-carbon chain single-bonded together with continuation bonds going through square brackets, and an 'n' subscript. One carbon has two $-\text{H}$ groups, and the other has one $-\text{H}$ and one $-\text{CH}_2\text{CH}_3$ (or $-\text{C}_2\text{H}_5$) side group.

(ii) Landfill: Addition polymers are inert and non-biodegradable, meaning they remain in the environment for hundreds of years and take up scarce space.
Combustion: Burning polymers produces greenhouse gases like carbon dioxide which contribute to global warming, and can release toxic gases such as carbon monoxide or hydrogen chloride.

(c) (i) The balanced equation is:
$\text{C}_4\text{H}_8 + 6\text{O}_2 \rightarrow 4\text{CO}_2 + 4\text{H}_2\text{O}$

(ii) Step-by-step calculation:
1. Find moles of $\text{C}_4\text{H}_8$:
$\text{moles} = \frac{\text{mass}}{M_r} = \frac{1.40}{56.0} = 0.025\text{ mol}$

2. Determine moles of $\text{CO}_2$ from the molar ratio (1 : 4):
$\text{moles of } \text{CO}_2 = 0.025 \times 4 = 0.10\text{ mol}$

3. Calculate the volume in $\text{dm}^3$ and convert to $\text{cm}^3$:
$\text{Volume in } \text{dm}^3 = 0.10\text{ mol} \times 24.0\text{ dm}^3/\text{mol} = 2.40\text{ dm}^3$
$\text{Volume in } \text{cm}^3 = 2.40 \times 1000 = 2400\text{ cm}^3$

(d) Test:
- Add bromine water (bromine solution) to both compounds.
- Observation with butane: The solution remains orange/brown.
- Observation with but-1-ene: The solution turns colourless / is decolourised.

Marking scheme

Part (a):
- (i) 1 mark for but-1-ene, 1 mark for but-2-ene (or 2-methylpropene).
- (ii) 1 mark for the correct fully displayed formula of but-1-ene; 1 mark for the correct fully displayed formula of but-2-ene / 2-methylpropene.

Part (b):
- (i) 1 mark for the correct backbone with a single C-C bond and correct substituent groups ($-\text{H}$ and $-\text{C}_2\text{H}_5$); 1 mark for square brackets, continuation bonds, and 'n'.
- (ii) 1 mark for identifying landfill issue (inert/non-biodegradable); 1 mark for identifying combustion issue (greenhouse gas emissions/toxic gas release).

Part (c):
- (i) 1 mark for correct reactants and products; 1 mark for correct balancing.
- (ii) 1 mark for calculating $0.025\text{ mol}$ of $\text{C}_4\text{H}_8$; 1 mark for using the correct stoichiometry to get $0.10\text{ mol}$ of $\text{CO}_2$ (allow ECF from incorrect equation); 1 mark for final volume of $2400\text{ cm}^3$ (accept $2.4\text{ dm}^3$ only if units are clearly stated, but deduct 1 mark if conversion to $\text{cm}^3$ is missing when specifically asked).

Part (d):
- 1 mark for adding bromine water/bromine solution.
- 1 mark for both correct observations (butane stays orange/brown AND but-1-ene decolourises/goes colourless).

Question 10 · written

15 marks

Show answer & marking scheme

Worked solution

Marking scheme

Question 11 · structured

11 marks

A student investigates the rate of reaction between calcium carbonate (marble chips) and dilute hydrochloric acid.

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

(a) State the name of the gas produced in this reaction and describe a chemical test, including the observation, to confirm its identity. [3]

(b) The student measures the volume of gas produced over time. Explain, in terms of collision theory, how increasing the temperature of the acid increases the rate of reaction. [4]

(c) The student repeats the experiment under the same conditions but uses the same mass of calcium carbonate powder instead of large marble chips.

(i) State and explain the effect of this change on the initial rate of reaction. [3]

(ii) State the effect, if any, on the final volume of gas collected. [1]

Show answer & marking scheme

Worked solution

(a) The gas produced is carbon dioxide. To test for its presence, bubble the gas through limewater (aqueous calcium hydroxide). The limewater will turn cloudy or milky due to the formation of an insoluble white precipitate of calcium carbonate.

(b) Increasing the temperature increases the thermal energy of the particles, which is converted to kinetic energy. As a result, the reactant particles move faster, leading to more frequent collisions per unit time. Furthermore, a much higher proportion of the colliding particles now possess energy that is greater than or equal to the activation energy, which significantly increases the rate of successful collisions.

(c) (i) Using powder instead of large chips increases the rate of reaction. This is because the powder has a much larger surface area to volume ratio, meaning more reactant particles are exposed and available to collide at any given time. This increases the frequency of collisions between the calcium carbonate and the acid particles.
(ii) There is no change to the final volume of gas collected because the limiting reactant and the quantities/masses used remain identical, so the same total moles of product are formed.

Marking scheme

(a)
- M1: Carbon dioxide (1)
- M2: Bubble the gas into limewater / calcium hydroxide solution (1) [Reject: Litmus/indicator tests]
- M3: Limewater turns cloudy / milky / forms a white precipitate (1)

(b)
- M1: Particles gain kinetic energy / move faster (1)
- M2: More frequent collisions / more collisions per unit time (1) [Accept: collisions occur more often]
- M3: More particles have energy greater than or equal to the activation energy (1)
- M4: More successful collisions per unit time / higher proportion of successful collisions (1) [Do not accept 'more successful collisions' on its own without a time reference or reference to proportion]

(c)(i)
- M1: Rate increases / reaction is faster (1)
- M2: Larger surface area (to volume ratio) (1)
- M3: More frequent collisions / more collisions per unit time (1)

(c)(ii)
- M1: No change / stays the same (1)

Question 12 · structured

11 marks

A student investigates the displacement reaction between zinc powder and copper(II) sulfate solution. The equation for the reaction is: $ \text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)} $. The student uses this method: 1. Pour $ 50.0\text{ cm}^3 $ of $ 0.200\text{ mol/dm}^3 $ copper(II) sulfate solution into a polystyrene cup. 2. Measure the initial temperature of the solution. 3. Add an excess of zinc powder and stir the mixture. 4. Measure the maximum temperature reached. Results: Initial temperature of solution = $ 19.5\text{ }^\circ\text{C} $, Maximum temperature of solution = $ 31.8\text{ }^\circ\text{C} $. (a) State why a polystyrene cup is used instead of a glass beaker. [1 mark] (b) Show that the heat energy change (q) in this reaction is approximately $ 2.57\text{ kJ} $. Assume the density of the solution is $ 1.00\text{ g/cm}^3 $ and its specific heat capacity is $ 4.18\text{ J/g/}^\circ\text{C} $. [3 marks] (c) Calculate the number of moles of copper(II) sulfate in $ 50.0\text{ cm}^3 $ of $ 0.200\text{ mol/dm}^3 $ solution. [2 marks] (d) Calculate the enthalpy change ($ \Delta H $) for the reaction, in $ \text{kJ/mol} $. Include a sign in your answer. [3 marks] (e) State two reasons why the student's experimental value for $ \Delta H $ is less exothermic than the accepted data book value. [2 marks]

Show answer & marking scheme

Worked solution

(a) Polystyrene is a good thermal insulator, which reduces heat loss to the surroundings. (b) Temperature change $ \Delta T = 31.8 - 19.5 = 12.3\text{ }^\circ\text{C} $. Heat energy change $ q = m \times c \times \Delta T = 50.0\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 12.3\text{ }^\circ\text{C} = 2570.7\text{ J} $. In kilojoules, $ q = \frac{2570.7}{1000} = 2.5707\text{ kJ} $, which is approximately $ 2.57\text{ kJ} $. (c) $ \text{Moles} = \text{volume in dm}^3 \times \text{concentration} = \frac{50.0}{1000} \times 0.200 = 0.0100\text{ mol} $. (d) $ \Delta H = -\frac{q}{\text{moles}} = -\frac{2.5707\text{ kJ}}{0.0100\text{ mol}} = -257.07\text{ kJ/mol} $ (or $ -257\text{ kJ/mol} $ using the rounded value of $ 2.57\text{ kJ} $). Since the reaction is exothermic (temperature increased), the sign is negative: $ -257\text{ kJ/mol} $. (e) Two reasons: 1. Heat lost to the surroundings (air or container). 2. Heat absorbed by the polystyrene cup and the thermometer itself.

Marking scheme

(a) 1 mark: Polystyrene is a good thermal insulator / reduces heat loss. (b) 3 marks: 1 mark for calculating temperature change ($ 12.3\text{ }^\circ\text{C} $); 1 mark for correct substitution into formula ($ 50.0 \times 4.18 \times 12.3 $); 1 mark for obtaining $ 2571\text{ J} $ or $ 2.57\text{ kJ} $. (c) 2 marks: 1 mark for dividing volume by 1000 ($ 0.0500\text{ dm}^3 $); 1 mark for final moles value of $ 0.0100\text{ mol} $. (d) 3 marks: 1 mark for dividing heat energy by moles; 1 mark for the correct magnitude ($ 257 $ or $ 260 $); 1 mark for the negative sign. Accept values from $ -257 $ to $ -260 $ inclusive. (e) 2 marks: 1 mark for each valid reason, e.g., heat loss to surroundings (1), heat absorbed by cup/thermometer (1), incomplete reaction (1).

Paper 2C

Answer all questions. Show all steps in calculations. Calculators and rulers are permitted.

8 Question · 75 marks

Question 1 · structured

7 marks

An element, X, has an atomic number of 15.

(a) State the electronic configuration of an atom of X. (1)

(b) Explain how this electronic configuration shows that element X is in Group 5 and Period 3 of the Periodic Table. (2)

(c) Argon is another element in Period 3. Explain, in terms of electronic configuration, why argon is very unreactive. (2)

(d) Magnesium (Group 2) reacts with element X to form an ionic compound. Deduce the formula of this compound. (2)

Show answer & marking scheme

Worked solution

**(a)**
An atom with an atomic number of 15 has 15 protons, and therefore 15 electrons.
- The first shell can hold up to 2 electrons.
- The second shell can hold up to 8 electrons.
- The remaining 5 electrons go into the third shell.
So, the electronic configuration is **2.8.5**.

**(b)**
- The number of outer shell (valency) electrons corresponds to the Group number. Since there are 5 electrons in the outer shell, it is in **Group 5**.
- The number of occupied electron shells corresponds to the Period number. Since there are 3 shells containing electrons, it is in **Period 3**.

**(c)**
Argon has an atomic number of 18, giving it an electronic configuration of 2.8.8. It has a **full outer shell** (8 electrons). Because its outer shell is complete, it is extremely stable and does not need to lose, gain, or share electrons to react.

**(d)**
- Magnesium is in Group 2, so it loses 2 electrons to form a $ \text{Mg}^{2+} $ ion.
- Element X is in Group 5, so it gains 3 electrons to form an $ \text{X}^{3-} $ (or $ \text{P}^{3-} $) ion.
- To balance the charges (lowest common multiple of 2 and 3 is 6):
- Three $ \text{Mg}^{2+} $ ions are needed: $ 3 \times (+2) = +6 $
- Two $ \text{X}^{3-} $ ions are needed: $ 2 \times (-3) = -6 $
- This gives the formula **$ \text{Mg}_3\text{X}_2 $** (or **$ \text{Mg}_3\text{P}_2 $** if identified as phosphorus).

Marking scheme

**(a)**
- **M1**: 2.8.5 (1)

**(b)**
- **M1**: (Group 5 because) there are 5 electrons in the outer shell (1)
- **M2**: (Period 3 because) there are 3 shells containing electrons / 3 energy levels (1)

**(c)**
- **M1**: Argon has a full outer shell (of electrons) / 8 electrons in outer shell / electronic configuration 2.8.8 (1)
- **M2**: It does not need to / cannot easily gain, lose, or share electrons (1)

**(d)**
- **M1**: Identifying the charges on the ions as $ \text{Mg}^{2+} $ and $ \text{X}^{3-} $ (or $ \text{P}^{3-} $) (1)
- **M2**: $ \text{Mg}_3\text{X}_2 $ / $ \text{Mg}_3\text{P}_2 $ (1)
*(Accept any order of the elements, but metals should ideally be written first. Reject formulas with charges left in the final answer, e.g., $ \text{Mg}^{2+}_3\text{X}^{3-}_2 $ for M2)*

Question 2 · descriptive

5 marks

A marine researcher investigates methods to prevent the steel (iron) hull of a ship from rusting. She attaches blocks of zinc to some parts of the hull and blocks of copper to other parts. Explain how attaching blocks of zinc protects the iron from rusting, and why copper blocks are not suitable for this purpose.

Show answer & marking scheme

Worked solution

Zinc is more reactive than iron, which means zinc atoms lose electrons more readily than iron atoms to form ions: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-$. Therefore, zinc reacts and corrodes instead of the iron. This process is called sacrificial protection. Because copper is less reactive than iron, it does not lose electrons as easily as iron. Thus, copper cannot sacrificially protect iron, and the iron will continue to rust (and may even rust faster) if copper is attached.

Marking scheme

M1: State that zinc is more reactive than iron (1 mark) M2: Explain that zinc loses electrons / oxidises more readily than iron (1 mark) M3: Explain that zinc corrodes or reacts instead of iron / acts sacrificially (1 mark) M4: State that copper is less reactive than iron (1 mark) M5: Explain that copper cannot protect iron / iron will still lose electrons and rust (1 mark)

Question 3 · descriptive

5 marks

Show answer & marking scheme

Worked solution

Marking scheme

Question 4 · structured

11 marks

This question is about the industrial separation and processing of crude oil. (a) Crude oil is a mixture of hydrocarbons. Explain how crude oil is separated into fractions in a fractional distillation column. [4 marks] (b) Long-chain alkanes obtained from fractional distillation are cracked to produce shorter-chain alkanes and alkenes. (i) State the conditions needed for catalytic cracking in industry. [2 marks] (ii) Write a balanced chemical equation for the cracking of decane, $C_{10}H_{22}$, to produce oct-1-ene, $C_8H_{16}$, and another hydrocarbon. [1 mark] (iii) Explain why the demand for shorter-chain alkanes is higher than the demand for long-chain alkanes, referring to their uses. [2 marks] (c) Describe a chemical test to distinguish between the two products of the cracking reaction in (b)(ii). State the observations for both products. [2 marks]

Show answer & marking scheme

Worked solution

(a) Crude oil is heated until it vaporises. The vapour enters a fractionating column which is hot at the bottom and cooler at the top. The different hydrocarbons rise up the column and condense when they reach their respective boiling points, allowing them to be collected as separate fractions. (b)(i) Catalytic cracking requires a high temperature of 600 to 700 degrees Celsius and a catalyst of silica or alumina. (ii) $C_{10}H_{22} \rightarrow C_8H_{16} + C_2H_6$ (iii) Shorter-chain alkanes are highly flammable and burn cleanly, making them highly effective fuels (like petrol) for which there is high demand, whereas longer-chain alkanes are viscous, less flammable, and less useful as fuels. (c) Add bromine water to both products. The alkene (oct-1-ene) will decolourise the bromine water from orange/brown to colourless, while the alkane (ethane) will show no change and the solution remains orange/brown.

Marking scheme

(a) M1: Crude oil is heated / vaporised. M2: There is a temperature gradient in the column (hotter at the bottom, cooler at the top). M3: Vapours rise and condense at their boiling points. M4: Fractions with lower boiling points condense higher up / fractions with higher boiling points condense lower down. (b)(i) M1: Temperature of 600 - 700 degrees Celsius (accept 500 - 750 degrees Celsius). M2: Silica / alumina catalyst (accept zeolite / aluminosilicate). (b)(ii) M1: Correct balanced equation: $C_{10}H_{22} \rightarrow C_8H_{16} + C_2H_6$. (b)(iii) M1: Shorter-chain alkanes make better/more flammable fuels (or specific fuel use such as petrol). M2: Long-chain alkanes are less useful / there is a surplus of long-chain fractions. (c) M1: Add bromine water (accept bromine solution). M2: Oct-1-ene decolourises the solution / turns from orange to colourless AND ethane does not change color / stays orange.

Question 5 · subjective

10 marks

A student investigates how the solubility of Salt Z in water changes with temperature. The table shows the student's results:

Temperature (°C)Solubility of Salt Z (g per 100 g of water)1053011502170379059

(a) Describe a practical method the student could use to determine the solubility of Salt Z at 50 °C. [4]

(b) (i) Calculate the mass of Salt Z that crystallises when a saturated solution of Salt Z in 150 g of water is cooled from 70 °C to 10 °C. [3]

(ii) A saturated solution of Salt Z with a total mass of 318 g is prepared at 90 °C. Calculate the mass of Salt Z that will precipitate out of the solution when it is cooled to 30 °C. [3]

Show answer & marking scheme

Worked solution

Part (a)

1. Add excess Salt Z to a beaker of water at 50 °C and stir thoroughly to ensure the solution is saturated (with visible undissolved solid remaining).
2. Filter the mixture to remove all undissolved solid, collecting only the clear saturated solution.
3. Pour a sample of the solution into a pre-weighed evaporating basin and weigh the basin plus solution to determine the starting mass.
4. Heat the basin to evaporate all the water until only dry crystals remain, then re-weigh to determine the mass of the dry salt.
5. Calculate the mass of water evaporated (mass of solution - mass of dry salt) and use the formula: $\text{Solubility} = \frac{\text{mass of salt}}{\text{mass of water}} \times 100$ to get the final value.

Part (b)(i)

At 70 °C, solubility = 37 g per 100 g of water.
At 10 °C, solubility = 5 g per 100 g of water.
Difference in solubility per 100 g of water = $37 - 5 = 32\text{ g}$.
For 150 g of water, the mass that crystallises out is: $32 \times \frac{150}{100} = 48\text{ g}$.

Part (b)(ii)

At 90 °C, solubility = 59 g per 100 g of water.
This means a standard saturated solution has a total mass of $100\text{ g (water)} + 59\text{ g (salt)} = 159\text{ g}$.
The actual solution mass is 318 g, which is exactly twice the standard scale ($\frac{318}{159} = 2$).
Therefore, the solution contains $100 \times 2 = 200\text{ g of water}$ and $59 \times 2 = 118\text{ g of Salt Z}$.
At 30 °C, solubility = 11 g per 100 g of water.
In 200 g of water at 30 °C, the mass of salt that can remain dissolved is: $11 \times 2 = 22\text{ g}$.
Therefore, the mass of Salt Z that precipitates out is: $118 - 22 = 96\text{ g}$.
(Alternative calculation: Difference in solubility = $59 - 11 = 48\text{ g per 100 g water}$. For 200 g of water, mass precipitated = $48 \times 2 = 96\text{ g}$).

Marking scheme

Part (a) [4 marks]
- M1: Stir excess Salt Z in water at 50 °C (to ensure saturation) [1]
- M2: Filter / decant the solution to remove undissolved solid [1]
- M3: Weigh the saturated solution in a pre-weighed evaporating basin, evaporate to dryness, and re-weigh to find the mass of dry salt [1]
- M4: Calculate the mass of water evaporated (solution mass - dry salt mass) and use $\text{Solubility} = \frac{\text{mass of salt}}{\text{mass of water}} \times 100$ [1]

Part (b)(i) [3 marks]
- M1: Correct subtraction of solubilities: $37 - 5 = 32\text{ g}$ [1]
- M2: Correct scaling: $32 \times 1.5$ (or equivalent) [1]
- M3: $48\text{ g}$ (correct final answer with no working scores 3 marks) [1]

Part (b)(ii) [3 marks]
- M1: Calculate total mass of standard saturated solution at 90 °C: $100 + 59 = 159\text{ g}$ OR deduce the actual mass of water is 200 g [1]
- M2: Show scaling factor of 2 (e.g. $\frac{318}{159} = 2$) OR show subtraction of solubilities: $59 - 11 = 48\text{ g}$ [1]
- M3: $96\text{ g}$ (correct final answer with no working scores 3 marks) [1]

Question 6 · subjective

10 marks

This question is about alcohols, carboxylic acids, and esters.

(a) Alcohols and carboxylic acids both belong to different homologous series. State two characteristics of a homologous series. (2)

(b) Ethanol reacts with butanoic acid in the presence of an acid catalyst to form an ester.

(i) State the name of the acid catalyst used in this reaction. (1)

(ii) Draw the displayed formula of the ester produced, showing all atoms and all bonds, and state its IUPAC name. (3)

(c) A student prepares ethyl ethanoate by heating a mixture of ethanol and ethanoic acid with a catalyst.

The student starts with a reaction mixture that has a theoretical yield of $22.0\text{ g}$ of ethyl ethanoate. After purification, the student collects $14.3\text{ g}$ of pure ethyl ethanoate.

(i) Calculate the percentage yield of ethyl ethanoate in this experiment. (2)

(ii) State one physical property of esters that makes them suitable for use in perfumes. (1)

(iii) State one other use of esters, apart from in perfumes. (1)

Show answer & marking scheme

Worked solution

### Part (a)
A homologous series is a family of organic compounds that share key characteristics:
* They have the **same general formula** (e.g., $\text{C}_n\text{H}_{2n+1}\text{OH}$ for alcohols).
* They exhibit **similar chemical properties** because they contain the same functional group.
* There is a **gradual trend in physical properties** (e.g., boiling point increases as molecular mass increases).
* Consecutive members differ from each other by a $\text{-CH}_2-$ unit.

### Part (b)
(i) The catalyst used in esterification reactions is **concentrated sulfuric acid** ($\text{H}_2\text{SO}_4$).

(ii) The reaction between ethanol and butanoic acid produces the ester **ethyl butanoate** and water.
* *Alcohol part:* Ethanol provides the ethyl group ($-\text{CH}_2\text{CH}_3$).
* *Carboxylic acid part:* Butanoic acid provides the butanoate group ($\text{CH}_3\text{CH}_2\text{CH}_2\text{COO}-$).
* *Displayed Formula:* All atoms (C, H, O) and all covalent bonds (including all single $\text{C-H}$, $\text{C-C}$, $\text{C-O}$ bonds and the double $\text{C=O}$ bond) must be explicitly shown:
```
H H H O H H
| | | // | |
H - C - C - C - C - O - C - C - H
| | | | |
H H H H H
```
* *Name:* **ethyl butanoate**.

### Part (c)
(i) To calculate the percentage yield:
\[\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\]
\[\text{Percentage Yield} = \frac{14.3\text{ g}}{22.0\text{ g}} \times 100 = 65.0\%\]

(ii) Esters are **volatile** (vaporise easily to carry the scent) and have a **distinctive sweet, fruity smell**, making them highly suitable for perfumes.

(iii) Esters are widely used as **solvents** (e.g., in nail varnish removers, glues, and paints) or as artificial **food flavourings**.

Marking scheme

**(a)** (Maximum 2 marks)
* **M1:** same general formula (1)
* **M2:** similar chemical properties OR trend in physical properties OR consecutive members differ by a $\text{CH}_2$ group (1)

**(b)(i)** (1 mark)
* **M1:** Concentrated sulfuric acid / conc. $\text{H}_2\text{SO}_4$ (1)
* *Reject:* dilute sulfuric acid / sulfuric acid without 'concentrated' or 'conc.' / other acids.

**(b)(ii)** (3 marks)
* **M1:** Correct IUPAC name: **ethyl butanoate** (1) *(accept ethylbutyrate)*
* **M2:** Displayed formula of butyl/acid part: 4 carbons in a chain with single bonds, $\text{C=O}$ double bond, and single $\text{C-O}$ bond (1)
* **M3:** Displayed formula of ethyl/alcohol part: 2 carbons in a chain attached to the single-bonded oxygen, with all hydrogen atoms and all bonds shown (1)
* *Note: Deduct 1 mark from the drawing score if any single bond is missing (e.g., O-H is drawn as OH, or C-H bonds are condensed as $\text{CH}_3$).*

**(c)(i)** (2 marks)
* **M1:** Correct substitution: $\frac{14.3}{22.0} \times 100$ (1)
* **M2:** Correct evaluation: $65\%$ / $65.0\%$ (1)

**(c)(ii)** (1 mark)
* **M1:** Sweet / fruity smell OR volatile / easily vaporised (1)

**(c)(iii)** (1 mark)
* **M1:** (Artificial) food flavouring(s) OR solvent(s) (1)
* *Accept:* named household solvent use such as nail polish remover.

Question 7 · structured

14 marks

This question is about salt preparation and electrolysis.

(a) A student wants to prepare a pure, dry sample of the insoluble salt, lead(II) sulfate ($\text{PbSO}_4$).

(i) Name two soluble substances that the student could react together to prepare lead(II) sulfate. (2)

(ii) Describe a practical method the student should use to obtain a pure, dry sample of lead(II) sulfate from the mixture of these two solutions. (5)

(b) The student also investigates electrolysis.

(i) During the electrolysis of concentrated aqueous sodium chloride, a gas is produced at the anode. Name this gas and write an ionic half-equation for its formation. (3)

(ii) During the electrolysis of molten lead(II) bromide ($\text{PbBr}_2$), lead is formed at the cathode. Write an ionic half-equation for the formation of lead at this electrode. State and explain whether this process is oxidation or reduction. (4)

Show answer & marking scheme

Worked solution

**Part (a)(i)**
To prepare an insoluble salt, we need to mix two soluble salts (a source of the cation and a source of the anion). Lead(II) nitrate is the only common soluble lead salt, and sodium sulfate (or potassium sulfate/ammonium sulfate/dilute sulfuric acid) is a suitable soluble sulfate.

**Part (a)(ii)**
1. Filter the mixture to separate the insoluble lead(II) sulfate precipitate (residue) from the solution (filtrate).
2. Wash the residue remaining in the filter paper with distilled water to remove any remaining soluble impurities.
3. Dry the solid residue using filter paper, in a warm oven, or in a desiccator. Do not use direct strong heat from a Bunsen burner as it might decompose the salt.

**Part (b)(i)**
During the electrolysis of aqueous sodium chloride, chloride ions ($\text{Cl}^-$) are attracted to the anode (positive electrode), where they lose electrons to form chlorine gas ($\text{Cl}_2$).
The half-equation is: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-$

**Part (b)(ii)**
At the cathode (negative electrode), lead(II) ions ($\text{Pb}^{2+}$) gain electrons to form lead metal atoms ($\text{Pb}$).
The half-equation is: $\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}$
This process is reduction because oxidation is loss of electrons, and reduction is gain of electrons (OIL RIG). Since $\text{Pb}^{2+}$ ions gain electrons, they are reduced.

Marking scheme

**Part (a)(i) [2 marks]**
* **M1:** Lead(II) nitrate (1)
* **M2:** Sodium sulfate / potassium sulfate / ammonium sulfate / dilute sulfuric acid (1)
*(Accept: correct chemical formulae, e.g., $\text{Pb(NO}_3$_2\), $\text{Na}_2\text{SO}_4$. Reject: insoluble sulfates)*

**Part (a)(ii) [5 marks]**
* **M1:** Filter the mixture / use filtration (1)
* **M2:** Wash the residue / precipitate / lead(II) sulfate (1)
* **M3:** with distilled water / deionised water (1)
* **M4:** Dry the solid / residue (1)
* **M5:** using filter paper / in a warm oven / in a desiccator (1)
*(Reject: 'heat with a Bunsen burner' for M5)*

**Part (b)(i) [3 marks]**
* **M1:** Chlorine / $\text{Cl}_2$ (1) *(Reject: chloride)*
* **M2:** $2\text{Cl}^- \rightarrow \text{Cl}_2$ (1) *(correct reactants and products)*
* **M3:** $+ 2\text{e}^-$ on the right-hand side OR $- 2\text{e}^-$ on the left-hand side (1) *(correct balancing)*

**Part (b)(ii) [4 marks]**
* **M1:** $\text{Pb}^{2+} \rightarrow \text{Pb}$ (1) *(correct reactants and products)*
* **M2:** $\text{Pb}^{2+} + 2\text{e}^- \rightarrow \text{Pb}$ (1) *(fully balanced half-equation)*
* **M3:** Reduction (1)
* **M4:** (Lead ions) gain electrons OR decrease in oxidation state (from +2 to 0) (1)

Question 8 · structured

13 marks

This question is about hydrogen chloride and covalent bonding.

(a) State what is meant by a covalent bond. [2 marks]

(b) Hydrogen chloride gas dissolves in water and in methylbenzene (an organic solvent). Explain why the solution of hydrogen chloride in water is acidic, while the solution of hydrogen chloride in methylbenzene is not acidic. [4 marks]

(c) Explain, in terms of bond breaking and bond making, why the reaction between hydrogen and chlorine to form hydrogen chloride is exothermic. [4 marks]

(d) Use the bond energies in the table to calculate the enthalpy change ($\Delta H$) in kJ/mol for the reaction:
$H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$

| Bond | Bond energy (kJ/mol) |
|---|---|
| H-H | 436 |
| Cl-Cl | 243 |
| H-Cl | 432 | [3 marks]

Show answer & marking scheme

Worked solution

Step-by-step solution:

**(a)** A covalent bond is defined as the strong electrostatic attraction between a shared pair of electrons and the nuclei of the two bonded atoms.

**(b)** Water is a polar solvent that causes hydrogen chloride molecules to dissociate (ionise) into hydrogen ions ($H^+$) and chloride ions ($Cl^-$). The presence of free $H^+$ ions makes the solution acidic. Methylbenzene is a non-polar organic solvent; hydrogen chloride dissolves without dissociating and remains as neutral covalent molecules. Because there are no free $H^+$ ions, the solution is not acidic.

**(c)** Energetics explanation:
1. Breaking the bonds in the reactants (H-H and Cl-Cl) is an endothermic process because it requires energy.
2. Forming the bonds in the product (H-Cl) is an exothermic process because it releases energy.
3. Since the energy released during the formation of the two H-Cl bonds is greater than the total energy absorbed to break the H-H and Cl-Cl bonds, the reaction is overall exothermic.

**(d)** Enthalpy calculation:
1. Energy absorbed to break reactant bonds = $436\text{ (H-H)} + 243\text{ (Cl-Cl)} = 679\text{ kJ/mol}$.
2. Energy released when making product bonds = $2 \times 432\text{ (H-Cl)} = 864\text{ kJ/mol}$.
3. $\Delta H = \text{Energy absorbed} - \text{Energy released} = 679 - 864 = -185\text{ kJ/mol}$.

Marking scheme

**Part (a)**
- **M1**: Shared pair of electrons (1 mark)
- **M2**: Electrostatic attraction between the electrons and the nuclei of the bonded atoms (1 mark)
*(Reject: attraction between ions)*

**Part (b)**
- **M1**: (In water) HCl dissociates / ionises (1 mark)
- **M2**: to form hydrogen ions / $H^+$ ions (1 mark)
- **M3**: (In methylbenzene) HCl does not dissociate / remains as molecules (1 mark)
- **M4**: Acidity is caused by the presence of hydrogen ions / $H^+$ (1 mark)

**Part (c)**
- **M1**: Bond breaking is endothermic / requires energy (1 mark)
- **M2**: Bond making is exothermic / releases energy (1 mark)
- **M3**: More energy is released than is absorbed / taken in (1 mark)
- **M4**: Reference to specific bonds: energy released in making H-Cl bonds is greater than energy absorbed in breaking H-H and Cl-Cl bonds (1 mark)

**Part (d)**
- **M1**: Calculation of energy absorbed: $436 + 243 = 679\text{ (kJ/mol)}$ (1 mark)
- **M2**: Calculation of energy released: $2 \times 432 = 864\text{ (kJ/mol)}$ (1 mark)
- **M3**: Correct final answer with sign: $-185\text{ kJ/mol}$ (1 mark)
*(Note: Award 2 marks max for +185)*

Wondering how well you actually know this?

Thinka is an AI practice app for DSE students — unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.

Start Practising Free See pricing

Want more questions like this? Practice unlimited on Thinka — instant answers included.

Start Practising Free