Question

At ${27}^{\circ }\text{C}$, hydrogen is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas at the same temperature and pressure leaked through the same hole for 20 minute. After the effusion of the gases the mixture exerts a pressure of 9 atm. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 L, what is the molecular mass of the unknown gas (in g)?

Answer 1

Given,
$T=27{}^{\circ }\text{C}=300K$
$n_{H_2}=0.7mol$
$\text{Volume of the container}=3L$
Let $P_{H_2}$ and $P_{un}$ be the partial pressures of hydrogen and unknown gas respectively and $n$ be the number of moles of unknown gas.
$P_{H_2}=\frac{0.7}{3}\times 0.0821\times 300$
$P_{un}=\frac{w}{3}\times 0.0821\times 300$
adding both,
$P_{H_2}+P_{un}=9=\left(1/3\right)\times 0.0821\times 300(0.7+n)$
$n=0.3962$ mole
Applying law of diffusion,
$\text{Rate of diffusion}=\frac{\text{Mole of gas effused}}{\text{time taken}}$
$\text{So}{r}_{H_2}=\frac{0.7}{20},r_g=\frac{0.3962}{20}$
$\frac{r_{H_2}}{r_g}=\sqrt{\frac{M_g}{M_{H_2}}}$
$\frac{0.7/20}{0.3962/20}=\sqrt{\frac{M_g}{2}}$ or $M_g=6.24g$