Question

At $27{}^{\circ }C$, hydrogen is leaked through a tiny hole into a vessel for 20 $min$. Another unknown gas at the same temperature and pressure as that of hydrogen are leaked through the same hole for 20 min. After the effusion of the gases, the mixture exerts a pressure of 6 atm. The hydrogen content of the mixture is 0.7 mol.

If the volume of the container is 3 $L$, the molecular weight of the unknown gas is:

Answer 1

$PV=nRT$

or $6\times 3=n\times 0.0821\times 300$

or $n=0.7308$

$n_{H_2}=0.7$ (given)

$n_{gas}=0.7308-0.7=0.0308$

Applying Graham's law of diffusion, we get

$\frac{r_{H_2}}{r_{gas}}=\sqrt{\frac{M_{gas}}{M_{H_2}}}$

or $\frac{n_{H_2}}{n_{gas}}=\sqrt{\frac{M_{gas}}{M_{H_2}}}$

(because the time of diffusion is the same)

or $\frac{0.7}{0.0308}=\sqrt{\frac{M_{gas}}{2}}$

or $M_{gas}=1033.05\approx 1033$ g