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AQA GCSE Chemistry Smart practice A mixed round aimed at your weak spots
Unit 1
Atomic structure and the periodic table
Unit 2
Bonding, structure, and the properties of matter
Unit 3
Quantitative chemistry
Unit 4
Chemical changes
Unit 5
Energy changes
Unit 6
The rate and extent of chemical change
Unit 7
Organic chemistry
Unit 8
Chemical analysis
Unit 9
Chemistry of the atmosphere
Unit 10
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The proportions of different gases in the atmosphere

Practise 15 exam-style questions on The proportions of different gases in the atmosphere 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.

  • 15 questions
  • 1–6 marks each
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15 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

1 mark

Name the gas that makes up about one-fifth of the Earth's atmosphere today.

Model answer

Oxygen

Explanation

Oxygen makes up about one-fifth (approximately 20%, more precisely 21%) of the atmosphere. Nitrogen makes up most of the rest.

Mark scheme
  1. 1 mark

    oxygen

    allow O₂

Question 2

2 marks

Describe how the proportions of the main gases in the Earth's atmosphere have changed over the last 200 million years.

Model answer

For the last 200 million years, the proportions of the gases have stayed much the same as they are today: about four-fifths (80%) nitrogen and one-fifth (20%) oxygen, with small proportions of other gases such as carbon dioxide, water vapour and noble gases.

Explanation

The big changes in the atmosphere happened much earlier in Earth's history. Over the last 200 million years the composition has been fairly stable.

Mark scheme
  1. 1 mark

    the proportions have stayed much the same (as today)

    allow little / no significant change

  2. 1 mark

    about 80% nitrogen and about 20% oxygen

    allow four-fifths nitrogen and one-fifth oxygen

Question 3

3 marks

Air is about 80% nitrogen and 20% oxygen by volume. A student says: "There are about four times as many nitrogen molecules as oxygen molecules in air."

Explain why the student is correct.

Model answer

At the same temperature and pressure, equal volumes of any gases contain the same number of molecules. In air, nitrogen takes up 80% of the volume and oxygen 20%, so the volume of nitrogen is 80 ÷ 20 = 4 times that of oxygen. This means there are about four times as many nitrogen molecules as oxygen molecules.

Explanation

Because one mole of any gas occupies the same volume at the same temperature and pressure, percentage by volume is the same as percentage by number of molecules. This is why atmospheric composition is usually given by volume.

Mark scheme
  1. 1 mark

    equal volumes of gases (at the same temperature and pressure) contain the same number of molecules

    allow one mole of any gas occupies the same volume (24 dm³ at RTP)

  2. 1 mark

    the volume of nitrogen is four times the volume of oxygen (80 ÷ 20 = 4)

  3. 1 mark

    so there are four times as many nitrogen molecules / moles

Question 4

4 marks

A student uses two gas syringes to pass 100 cm³ of air backwards and forwards over heated copper. The copper reacts with the oxygen in the air to form copper oxide. The apparatus is then left to cool to room temperature. The final volume of gas is 79 cm³.

Calculate the percentage of oxygen in the air. Suggest why the gas is cooled before the final volume is read and why the air is passed over the copper many times.

Model answer

volume of oxygen removed = 100 − 79 = 21 cm³
percentage of oxygen = (21 ÷ 100) × 100 = 21%
The gas is cooled because hot gas expands, so reading the volume while hot would make the final volume too large; both volumes must be measured at the same temperature. The air is passed over the copper many times so that all of the oxygen has a chance to react.

Explanation

The decrease in volume equals the volume of oxygen removed. The result of 21% agrees with the accepted value of about one-fifth for oxygen in air.

Mark scheme
  1. 1 mark

    volume of oxygen = 100 − 79 = 21 cm³

    correct answer with no working scores full marks

  2. 1 mark

    percentage = 21 ÷ 100 × 100 = 21%

    allow ecf

  3. 1 mark

    hot gas has expanded / gas volume depends on temperature, so the reading would be too high

    allow so start and end volumes are measured at the same temperature

  4. 1 mark

    to make sure all the oxygen reacts with the copper

    allow until the volume stops changing

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