2022 H2 Chem P1 Q4 — Gases Solution

The answer is C (1.60 kg).

Total amount of gas: \(n = \frac{pV}{RT} = \frac{3.55 \times 10^7 \times 5 \times 10^{-3}}{8.31 \times (273 + 20)} = 72.9\) mol
(5 dm\(^3\) = \(5 \times 10^{-3}\) m\(^3\), and T = 293 K.)

Entonox is an equimolar mixture, so \(n(\text{N}_2\text{O}) = \frac{1}{2} \times 72.9 = 36.45\) mol.
\(M_\text{r}(\text{N}_2\text{O}) = 2(14.0) + 16.0 = 44.0\)
Mass of \(\text{N}_2\text{O}\) = \(36.45 \times 44.0 = 1604\) g = 1.60 kg

(Equivalently, the partial pressure of \(\text{N}_2\text{O}\) is \(\frac{1}{2} \times 3.55 \times 10^7 = 1.775 \times 10^7\) Pa, and \(m = \frac{pVM_\text{r}}{RT}\) gives the same 1604 g.)

• A is wrong: \(1.58 \times 10^{-3}\) kg comes from two slips: converting 5 dm\(^3\) to \(5 \times 10^{-6}\) m\(^3\) instead of \(5 \times 10^{-3}\) m\(^3\), and using 25 °C (298 K) instead of 20 °C (293 K): \(\frac{1.775 \times 10^7 \times 5 \times 10^{-6} \times 44.0}{8.31 \times 298} = 1.58\) g.
• B is wrong: \(1.60 \times 10^{-3}\) kg is 1000 times too small, which comes from converting 5 dm\(^3\) to \(5 \times 10^{-6}\) m\(^3\) instead of \(5 \times 10^{-3}\) m\(^3\).
• D is wrong: 3.21 kg is \(72.9 \times 44.0\) g, the mass if all the gas were \(\text{N}_2\text{O}\); only half of the moles are \(\text{N}_2\text{O}\).