Q.25. Equal masses of H2( g) and He(g) are enclosed in a container at constant temperature. The ratio of partial pressure of H2 to He is
A. 1:1
B. 1:2
C. 2:1
D. 1:4
Answer :- C. 2:1
Explanation :-
\text{Let the mass be } x. \mathrm{n}_{\mathrm{H}_2} = \frac{\mathrm{x} \, \mathrm{g}}{2 \, \mathrm{g} \, \mathrm{mol}^{-1}} = \frac{\mathrm{x}}{2} \mathrm{n}_{\mathrm{He}} = \frac{\mathrm{x} \, \mathrm{g}}{4 \, \mathrm{g} \, \mathrm{mol}^{-1}} = \frac{\mathrm{x}}{4} \mathrm{n}_{\text{Total}} = \frac{3 \mathrm{x}}{4} \text{Now,} x_{\mathrm{H}_2} = \frac{\mathrm{n}_{\mathrm{H}_2}}{\mathrm{n}_{\text{Total}}} = \frac{\mathrm{x} / 2}{3 \mathrm{x} / 4} = \frac{2}{3} x_{\mathrm{He}} = \frac{\mathrm{n}_{\mathrm{He}}}{\mathrm{n}_{\text{Total}}} = \frac{\mathrm{x} / 4}{3 \mathrm{x} / 4} = \frac{1}{3} \text{Now,} \mathrm{P}_{\mathrm{H2}} = x_{\mathrm{H}_2} \times \mathrm{P}_{\text{Total}} = \frac{2}{3} \times \mathrm{P} \mathrm{P}_{\mathrm{He}} = x_{\mathrm{He}} \times \mathrm{P}_{\text{Total}} = \frac{1}{3} \times \mathrm{P} \mathrm{P}_{\mathrm{H}_2} : \mathrm{P}_{\mathrm{He}} = \frac{2 / 3 \, \mathrm{P}}{1 / 3 \, \mathrm{P}} = 2 : 1