AQA GCSE Combined Science: Trilogy Chemistry Paper 1 (Higher), 2022: Question 2

8 marks · Standard Demand difficulty · Short Answer

Answer a series of questions about the electrolysis of potassium sulfate solution, including ionic formula, gas volumes, ratio explanation, uncertainty, and a concentration calculation.

Practise this question

Question

The question page is headed 'This question is about electrolysis' and shows a labelled diagram of an electrolysis setup for potassium sulfate solution. Two inverted 50 cm³ measuring cylinders collect gases above inert electrodes in a solution connected to a power supply; the left cylinder is labelled oxygen and the right hydrogen, with visible liquid levels indicating about 15 cm³ oxygen and 30 cm³ hydrogen. Below the figure are five parts: choosing the correct formula for potassium sulfate from four options, reading the gas volumes from the diagram, explaining how the gas volumes support a hypothesis about the H:O ratio in water, finding uncertainty from four oxygen volumes of 6, 9, 10 and 11 cm³ with mean 9 cm³, and calculating the mass of potassium sulfate needed to make 1.0 dm³ of solution from 0.86 g in 25 cm³.
Question text

02 This question is about electrolysis.

Figure 2 shows the apparatus used to investigate the electrolysis of

potassium sulfate solution.

Figure 2

02.1 Potassium sulfate contains K+ and SO 2– ions.

What is the formula of potassium sulfate?

[1 mark]

Tick ( ) one box.

KSO4

K2SO4

K(SO4)2

K2(SO4)2 7

02.2 What are the volumes of gases collected in the electrolysis experiment?

Use Figure 2.

[1 mark]

Volume of hydrogen = cm3

Volume of oxygen = cm3

02.3 A student made the following hypothesis:

‘The volumes of gases collected in this electrolysis experiment are in the same ratio

as hydrogen atoms to oxygen atoms in a water molecule.’

Explain how the volumes of gases collected in the experiment in Figure 2 support the

student’s hypothesis.

Use your answer to Question 02.2

[2 marks]

02.4 The experiment is repeated 4 times.

The volumes of oxygen collected in the 4 experiments are:

6 cm3 9 cm3 10 cm3 11 cm3

The mean volume of oxygen collected in the 4 experiments is 9 cm3

The measure of uncertainty is the range of a set of measurements about the mean.

What is the measure of uncertainty in the 4 experiments?

[1 mark]

Tick ( ) one box.

9 ± 1 cm3

*07* 9 ± 2 cm3

9 ± 3 cm3

02.5 The potassium sulfate solution has 0.86 g of potassium sulfate dissolved in

25 cm3 of water.

Calculate the mass of potassium sulfate needed to make 1.0 dm3 of solution.

[3 marks]

Mass = g

Mark scheme

Show the mark scheme The mark scheme is presented in tables for Questions 02.1 to 02.5 with columns for answers, extra information, marks, and AO/specification references. It gives K2SO4 for 02.1; 30 cm³ hydrogen and 15 cm³ oxygen for 02.2; for 02.3 it awards marks for stating the hydrogen to oxygen volume ratio is 2:1 and that this matches the hydrogen to oxygen atom ratio in H2O, with reference to the values from 02.2. For 02.4 the answer is 9 ± 3 cm³, and for 02.5 the calculation converts 25 cm³ to 0.025 dm³ and finds 34.4 g per dm³, with alternative equivalent methods allowed; total marks for Question 2 are 8.

Question 2

AO /

Question Answers Extra information Mark

Spec. Ref.

02.1 K2SO4 1 AO2

5.1.1.1

5.4.3.4

RPA9

AO /

Spec. Ref.

02.2 (volume of hydrogen) 30 (cm3) 1 AO2

and 5.4.3.4

(volume of oxygen) 15 (cm3) RPA9

AO /

Spec. Ref.

02.3 (because) the ratio of volume of 1 AO3

hydrogen : oxygen is 2 : 1 5.4.3.4

RPA9

(and this is the) same as the ratio of 1

hydrogen (atoms) : oxygen (atoms)

in (formula of) H2O

OR

(because) the ratio of volume of must relate to the volumes

hydrogen : oxygen is not 2 : 1 (1) given in question 02.2

(and this is) different to the ratio of

hydrogen (atoms) : oxygen (atoms)

in (formula of) H2O (1)

AO /

Spec. Ref.

02.4 9 ± 3 cm3 1 AO2

5.3.1.4

AO /

Question Answers Extra information Mark 9

Spec. Ref.

02.5 (conversion) AO2

25 3 5.3.2.5

( = ) 0.025 (dm ) 1

1000 5.4.3.4

(concentration =) allow correct use of incorrect / 1

0.86 no conversion

0.025

= 34.4 (g per dm3) allow 34 (g per dm3) 1

OR

(conversion)

1000

(1)

= 40 (1)

(40 × 0.86) allow correct use of incorrect /

= 34.4 (g per dm3) (1) no conversion

allow 34 (g per dm3)

OR

(concentration =)

0.86

(1)

= 0.0344 (1)

(conversion)

(0.0344 × 1000)

= 34.4 (g per dm3) (1) 3

allow 34 (g per dm )

Total Question 2 8

How to answer it

Electrolysis of Potassium Sulfate Solution

What this question tests

Recognising the formula of an ionic compound, reading values from a diagram, linking gas volumes to the water electrolysis ratio, using uncertainty as a range about the mean, and scaling up concentration using unit conversion from cm³ to dm³.

Question 02.1–02.5 Overview

Total marks: 8

This is a mostly straightforward calculation and recall question. The key marks come from accurate formula writing, reading the diagram carefully, and showing the working clearly in the calculation.

Part (a) 02.1 — Formula of potassium sulfate

What you needed to know

💡 Key knowledge

  • Potassium ions are K+ .
  • Sulfate ions are SO₄²− .
  • The total charge must balance to zero in the formula.

✅ Correct answer

K₂SO₄

1 mark for the correct neutral formula.

❌ Common errors

  • KSO₄ ignores the 2− charge on sulfate.
  • K(SO₄)₂ has the wrong ratio.
  • K₂(SO₄)₂ is not neutral.

🧠 Exam technique

When ionic charges are given, combine ions so the overall charge adds to zero. For a 2− ion like sulfate, you need two 1+ potassium ions.

Part (b) 02.2 — Volumes of gases collected

Reading the apparatus diagram

💡 Key knowledge

  • The diagram shows hydrogen collected at one electrode and oxygen at the other.
  • Use the scale on the measuring cylinders carefully.
  • The gas with the larger volume is hydrogen.

✅ Correct answers

  • Volume of hydrogen = 30 cm³
  • Volume of oxygen = 15 cm³

1 mark for both values correct.

🧠 Exam technique

Read from the bottom of the meniscus/level shown in the cylinder. Check the labelled gases and do not swap hydrogen and oxygen.

❌ Common errors

  • Mixing up which cylinder contains hydrogen and which contains oxygen.
  • Giving the difference between the volumes instead of the actual values.
  • Forgetting units: answers should be in cm³ .

Part (c) 02.3 — Explaining the ratio

How the volumes support the hypothesis

✅ Correct answer

The ratio of hydrogen : oxygen is 30 : 15 = 2 : 1, which is the same as the ratio of hydrogen atoms : oxygen atoms in H₂O.

2 marks: one for the 2:1 gas ratio, one for linking it to the atom ratio in water.

💡 Key knowledge

  • Water is H₂O , so it has 2 hydrogen atoms for every 1 oxygen atom.
  • During electrolysis, water breaks down into hydrogen and oxygen in this same proportion by volume.

❌ Common errors

  • Stating only the values without explaining the ratio.
  • Saying the ratio is 30:15 without simplifying it.
  • Comparing to the wrong formula or writing H2O without explaining the atom ratio.

🧠 Exam technique

To get both marks, always do two things: calculate the ratio from the data, then compare it to the atom ratio in water. The mark scheme also allowed the reverse idea: saying the ratio is not 2:1 and therefore does not match water, but this must still be linked to the measured volumes from 02.2.

Part (d) 02.4 — Uncertainty from repeated measurements

Mean and range about the mean

📐 Calculation

Measurements: 6 cm³, 9 cm³, 10 cm³, 11 cm³

  1. Mean is given as 9 cm³.
  2. Find the spread around the mean: lowest is 6, highest is 11.
  3. Difference from the mean:
    • 9 − 6 = 3 cm³
    • 11 − 9 = 2 cm³
  4. The larger deviation is 3 cm³.
  5. So the uncertainty is 9 ± 3 cm³.

✅ Correct answer

9 ± 3 cm³

1 mark for selecting the correct uncertainty.

❌ Common errors

  • Choosing ±2 because that is the distance from 9 to 11 only.
  • Using the range as 11 − 6 = 5 and then writing ±5.
  • Forgetting that uncertainty is the largest difference from the mean, not half the range here.

🧠 Exam technique

When the question says “range of a set of measurements about the mean,” look for the furthest measurement above or below the mean. The final answer should keep the mean value and add the uncertainty to it.

Part (e) 02.5 — Scaling up concentration

Mass needed to make 1.0 dm³ of solution

📐 Step-by-step calculation

  1. Find the concentration of the given solution:
    concentration = 0.86 ÷ 25 = 0.0344 g per cm³
  2. Convert to g per dm³ by multiplying by 1000:
    0.0344 × 1000 = 34.4 g per dm³
  3. Or convert the volume first:
    25 cm³ = 0.025 dm³
    0.86 ÷ 0.025 = 34.4 g per dm³
  4. For 1.0 dm³, the mass needed is 34.4 g.

✅ Correct answer

34.4 g

3 marks for correct method, correct conversion, and correct final answer.

💡 Key knowledge

  • 1 dm³ = 1000 cm³
  • 1.0 dm³ is a volume, but the question asks for the mass of solute in that volume.
  • Concentration in this question is in g per dm³ .

🧠 Exam technique

  • Show the conversion clearly: 25 ÷ 1000 = 0.025 dm³ .
  • Keep units attached to every step.
  • Round sensibly: 34.4 g is acceptable; the mark scheme also allowed 34 g .

❌ Common calculation traps

  • Using 25 as if it were already in dm³.
  • Forgetting to convert cm³ to dm³.
  • Dividing by 1000 instead of multiplying when converting g per cm³ to g per dm³.
  • Writing the answer without units.

✅ What a full-mark response looks like

25 cm³ = 0.025 dm³
concentration = 0.86 ÷ 0.025 = 34.4 g dm⁻³
mass needed for 1.0 dm³ = 34.4 g

Examiner insight: what separated strong answers

💡 Top-level responses

  • Used the correct formula from ion charges without hesitation.
  • Read the diagram accurately and wrote the gas volumes with units.
  • Linked the 2:1 gas ratio directly to H₂O.
  • Showed clear unit conversion in the concentration calculation.

❌ Where marks were likely lost

  • Part (c): giving the ratio but not explaining why it supports the hypothesis.
  • Part (d): choosing the wrong uncertainty from the list.
  • Part (e): using the wrong units or missing the conversion between cm³ and dm³.

🧠 Quick memory tips

  • Ion formula: balance charges.
  • Gas ratio in water electrolysis: hydrogen : oxygen = 2 : 1.
  • Uncertainty: use the furthest value from the mean.
  • Concentration: always check the units before calculating.

Topics

Chemistry · Required Practicals · C3: Quantitative Chemistry · C4: Chemical Changes · Chemistry Required Practicals

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Chemistry Paper 1 (Higher), 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.