Use of amount of substance in relation to volumes of gases (chemistry only) (HT only) - AQA GCSE Chemistry Exam Questions

Practise 22 exam-style questions on Use of amount of substance in relation to volumes of gases (chemistry only) (HT only) for AQA GCSE Chemistry. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

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Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

At room temperature and pressure, 1 mole of hydrogen gas and 1 mole of carbon dioxide gas both occupy 24 dm³, even though carbon dioxide molecules are much larger and heavier than hydrogen molecules.

Explain why.

Model answer

One mole of each gas contains the same number of molecules, 6.02 × 10²³. In a gas the molecules are very far apart, so the volume the gas takes up depends on the number of molecules rather than on their size or mass.

Explanation

Equal amounts in moles of gases occupy the same volume under the same conditions of temperature and pressure. Almost all of a gas's volume is empty space between the particles, so the size of each particle hardly matters; only the number of particles does.

Mark scheme

1 mark
1 mole of any gas contains the same number of molecules (6.02 × 10²³)
allow equal amounts in moles contain equal numbers of particles
1 mark
the molecules in a gas are very far apart, so the volume depends on the number of molecules, not on their size or mass
allow the size of the molecules is negligible compared with the space between them

Question 2

3 marks

Calculate the volume, in dm³, occupied by 8.8 g of carbon dioxide (CO₂) at room temperature and pressure.
Relative atomic masses (Ar): C = 12, O = 16
The volume of one mole of any gas at room temperature and pressure is 24 dm³.

Model answer

Mr of CO₂ = 12 + (2 × 16) = 44
amount = mass ÷ Mr = 8.8 ÷ 44 = 0.20 mol
volume = 0.20 × 24 = 4.8 dm³

Explanation

To get from a mass of gas to its volume, convert the mass to moles using the Mr, then multiply by the molar volume of 24 dm³. The same two steps work in reverse to find a mass from a volume.

Mark scheme

1 mark
Mr of CO₂ = 44
1 mark
amount = 8.8 ÷ 44 = 0.20 mol
allow ecf from an incorrect Mr
1 mark
volume = 0.20 × 24 = 4.8 (dm³)
allow ecf; correct answer with no working scores 3 marks

Question 3

3 marks

Zinc reacts with excess dilute hydrochloric acid:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)

A student needs to make 120 cm³ of hydrogen at room temperature and pressure.

Calculate the mass of zinc the student should use.
Relative atomic mass (Ar): Zn = 65
The volume of one mole of any gas at room temperature and pressure is 24 dm³.

Model answer

amount of H₂ = 120 ÷ 24 000 = 0.00500 mol
From the equation, 1 mol Zn gives 1 mol H₂, so amount of Zn = 0.00500 mol
mass of Zn = 0.00500 × 65 = 0.325 g

Explanation

Working backwards from a gas volume: convert the volume to moles, use the equation to find the moles of the reactant, then multiply by Ar to find the mass. Make sure the volume and molar volume are in the same units.

Mark scheme

1 mark
amount of H₂ = 0.120 ÷ 24 = 0.00500 mol
allow 120 ÷ 24 000
1 mark
amount of Zn = 0.00500 mol
1 : 1 ratio; allow ecf
1 mark
mass = 0.00500 × 65 = 0.325 (g)
allow 0.33; allow ecf; correct answer with no working scores 3 marks

Question 4

4 marks

Octane (C₈H₁₈) is a compound in petrol. It burns completely in oxygen:
2C₈H₁₈(l) + 25O₂(g) → 16CO₂(g) + 18H₂O(l)

Calculate the volume of carbon dioxide, in dm³, produced at room temperature and pressure when 57 g of octane burns completely.
Relative atomic masses (Ar): H = 1, C = 12
The volume of one mole of any gas at room temperature and pressure is 24 dm³.

Model answer

Mr of C₈H₁₈ = (8 × 12) + (18 × 1) = 114
amount of octane = 57 ÷ 114 = 0.50 mol
From the equation, 2 mol octane give 16 mol CO₂ (1 : 8), so amount of CO₂ = 0.50 × 8 = 4.0 mol
volume of CO₂ = 4.0 × 24 = 96 dm³

Explanation

Octane is a liquid, so its amount must be found from its mass, but carbon dioxide is a gas, so its volume can be found from moles × 24 dm³. Simplify the ratio 2 : 16 to 1 : 8 before using it.

Mark scheme

1 mark
Mr of C₈H₁₈ = 114
1 mark
amount of octane = 57 ÷ 114 = 0.50 mol
allow ecf from an incorrect Mr
1 mark
amount of CO₂ = 0.50 × 16 ÷ 2 = 4.0 mol
allow ecf
1 mark
volume = 4.0 × 24 = 96 (dm³)
allow ecf; correct answer with no working scores 4 marks

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