Q 11-12-092JEE MainJEE Main 2022 (27 Jun, Shift 1)Medium
A mixture of hydrogen and oxygen has volume $2000\ \text{cm}^3$, temperature $300$ K, pressure $100$ kPa and mass $0.76$ g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be [Take gas constant $R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$]
Answer: (B) $\dfrac{3}{1}$
Total moles:
$$n = \frac{PV}{RT} = \frac{(10^5)(2\times10^{-3})}{8.3\times300} = \frac{200}{2490} \approx 0.0803$$
Let $n_1$ = moles of $\text{H}_2$ (2 g/mol) and $n_2$ = moles of $\text{O}_2$ (32 g/mol):
$n_1 + n_2 = 0.0803$ and $2n_1 + 32n_2 = 0.76$.
Subtracting $2\times$ the first: $30n_2 = 0.76 - 0.1606 = 0.599$, so $n_2 \approx 0.020$ and $n_1 \approx 0.060$.
$$\frac{n_1}{n_2} = \frac{3}{1}$$
Solution by Sreeraj P, M.Sc Physics