Calculations involving gas volumes
4 marks · NormalThe table below shows the volume of different gases collected at rtp from several experiments.
| Experiment | Gas | Moles used | Volume collected (dm³) |
|---|---|---|---|
| 1 | H₂ | 0.50 | 12.0 |
| 2 | O₂ | 1.00 | 24.2 |
| 3 | CO₂ | 0.25 | 6.2 |
Identify which experiment gave a result closest to the expected molar volume at rtp. Show your working.
Suggest one reason why experiment 3's result is slightly higher than the expected value.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Experiment 2 (O₂); 24.2 ÷ 1.00 = 24.2 dm³ per mole (closest to 24 dm³) · must show correct division for each experiment, identification
(b) Collected air mixed in, not all CO₂ removed water vapour, or gas measured at higher than rtp temperature/pressure. · must show valid reason from possible sources of error
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Divides volume by moles for each experiment (e.g., 12.0/0.50=24, 24.2/1.00=24.2, 6.2/0.25=24.8)
A1 (a) · 1 mark · Identifies experiment 2 as closest to 24 dm³
B1 (b) · 1 mark · Gives a valid reason for slightly high result (e.g., slight contamination, not all air expelled)
B2 (b) · 1 mark · Elaboration or other valid source of error (e.g., imbalance at rtp, water vapour present)
Worked solution
a) Calculating molar volumes:
Experiment 1: 12.0 ÷ 0.50 = 24.0 dm³/mol
Experiment 2: 24.2 ÷ 1.00 = 24.2 dm³/mol
Experiment 3: 6.2 ÷ 0.25 = 24.8 dm³/mol
Experiment 2 is closest to the expected value of 24 dm³ per mole at rtp.
b) Reason: Slight error from not all air removed, water vapour in gas measured, or temperature higher than rtp.