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 $8.2\text{atm}$. The hydrogen content of the mixture is $0.7\text{mole}$. If the volume of the container is $3\text{L}$, what is the molecular mass of the unknown gas (in $\text{g}$)? Report your answer upto two decimal only.
(Take $R=0.082\text{L atm}{\text{mol}}^{-1}{\text{K}}^{-1}$)

Answer 1

Given,
$T=27{}^{\circ }\text{C}=300\text{K}$
$n_{H_2}=0.7\text{mol}$
$\text{Volume of the container}=3\text{L}$
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.
$P_{H_2}=\frac{0.7}{3}\times 0.082\times 300$
$P_{un}=\frac{n}{3}\times 0.082\times 300$
Adding both,
$P_{H_2}+P_{un}=8.2=\left(1/3\right)\times 0.082\times 300(0.7+n)$
$\Rightarrow n=0.3\text{mol}$
Applying law of diffusion,
$\text{Rate of diffusion}=\frac{\text{Mole of gas effused}}{\text{Time taken}}$
$\text{So},r_{H_2}=\frac{0.7}{20},r_{un}=\frac{0.3}{20}$
$\therefore \frac{r_{H_2}}{r_{un}}=\sqrt{\frac{M_g}{M_{H_2}}}$
$\Rightarrow \frac{0.7/20}{0.3/20}=\sqrt{\frac{M_g}{2}}$
$\Rightarrow M_g=10.88\text{g}$