Moles (HT only) - AQA GCSE Chemistry Exam Questions

Practise 28 exam-style questions on Moles (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.

28 questions
1–6 marks each
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20 questions, marked as you go

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

What is meant by the Avogadro constant? Give its value.

Model answer

The Avogadro constant is the number of atoms, molecules or ions in one mole of a substance. Its value is 6.02 × 10²³ per mole.

Explanation

One mole of any substance contains the same number of its stated particles, 6.02 × 10²³. Whether the particles are atoms, molecules or ions depends on the substance, so always state which particles you mean.

Mark scheme

1 mark
the number of atoms, molecules or ions (particles) in one mole of a substance
allow number of particles in one mole
1 mark
6.02 × 10²³ (per mole)
allow 6.0 × 10²³; ignore units

Question 2

2 marks

A sample of 0.20 mol of a gas has a mass of 6.4 g.

Calculate the relative formula mass (Mr) of the gas and use it to identify the gas from this list:

nitrogen (N₂), oxygen (O₂), carbon dioxide (CO₂), chlorine (Cl₂)

Relative atomic masses (Ar): C = 12, N = 14, O = 16, Cl = 35.5

Model answer

Mr = mass ÷ moles = 6.4 ÷ 0.20 = 32
The gas is oxygen, O₂ (Mr = 2 × 16 = 32).

Explanation

Rearranging moles = mass ÷ Mr gives Mr = mass ÷ moles. Then compare with the Mr of each gas: N₂ = 28, O₂ = 32, CO₂ = 44 and Cl₂ = 71.

Mark scheme

1 mark
Mr = 6.4 ÷ 0.20 = 32
correct answer with no working scores full marks
1 mark
oxygen / O₂
allow ecf from their Mr if it matches a gas in the list

Question 3

3 marks

A glass contains 180 cm³ of water. The density of water is 1.0 g/cm³, so the mass of the water is 180 g.

Calculate the amount of water in moles and the number of water molecules in the glass.

Relative atomic masses (Ar): H = 1, O = 16
The Avogadro constant is 6.02 × 10²³ per mole.

Model answer

Mr of H₂O = (2 × 1) + 16 = 18
moles = 180 ÷ 18 = 10 mol
number of molecules = 10 × 6.02 × 10²³ = 6.02 × 10²⁴

Explanation

A glass of water contains 10 moles, which is about six million million million million molecules. Multiplying 6.02 × 10²³ by 10 increases the power of ten by one.

Mark scheme

1 mark
Mr of H₂O = 18
correct answer with no working scores full marks
1 mark
moles = 180 ÷ 18 = 10 (mol)
allow ecf from Mr
1 mark
10 × 6.02 × 10²³ = 6.02 × 10²⁴ molecules
allow ecf; allow 60.2 × 10²³

Question 4

4 marks

Show that 4.6 g of sodium contains the same number of atoms as 3.2 g of oxygen gas (O₂).

Relative atomic masses (Ar): O = 16, Na = 23

Model answer

Sodium: moles of Na atoms = 4.6 ÷ 23 = 0.20 mol
Oxygen: Mr of O₂ = 32, so moles of O₂ = 3.2 ÷ 32 = 0.10 mol
Each O₂ molecule has 2 atoms, so moles of O atoms = 0.20 mol
Both samples contain 0.20 mol of atoms, so they contain the same number of atoms (1.204 × 10²³).

Explanation

Equal amounts in moles mean equal numbers of particles, so you do not need to multiply by the Avogadro constant to compare. Remember that oxygen gas is diatomic: one mole of O₂ contains two moles of oxygen atoms.

Mark scheme

1 mark
moles of Na = 4.6 ÷ 23 = 0.20 (mol)
1 mark
moles of O₂ = 3.2 ÷ 32 = 0.10 (mol)
1 mark
moles of O atoms = 0.10 × 2 = 0.20 (mol)
1 mark
both are 0.20 mol of atoms, so the same number of atoms
allow both 1.204 × 10²³ atoms

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