Showing that air contains about one-fifth oxygen
3 marks · NormalA sample of dry air has a total volume of 200 cm³. Calculate:
The volume of nitrogen in the sample.
The volume of oxygen in the sample.
The total volume of argon and carbon dioxide in the sample, using the approximate percentages in dry air.
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(a) 156 cm³ cm^3
(b) 42 cm³ cm^3
(c) 2 cm³ cm^3
Common mistakes: Mixing up nitrogen and oxygen percentages; Arithmetic errors
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Marking points
A1 (a) · 1 mark · Correct nitrogen volume: 78% of 200 = 156 cm³ (accept 156–157 cm³)
A2 (b) · 1 mark · Correct oxygen volume: 21% of 200 = 42 cm³ (accept 42 cm³)
A3 (c) · 1 mark · Sum of volumes for argon (0.9%) and CO₂ (0.04%): (0.009+0.0004)×200 ≈ 2 cm³ (accept 1.8–2.0 cm³)
Worked solution
a) Nitrogen: 78% of 200 = 0.78 × 200 = 156 cm³
b) Oxygen: 21% of 200 = 0.21 × 200 = 42 cm³
c) Argon: 0.9% of 200 = 0.009 × 200 = 1.8 cm³; Carbon dioxide: 0.04% of 200 = 0.0004 × 200 = 0.08 cm³; Total = 1.8 + 0.08 = 1.88 cm³ ≈ 2 cm³