States of Matter 1 Question 81
82. At $27^{\circ} \mathrm{C}$, hydrogen is leaked through a tiny hole into a vessel for $20 \mathrm{~min}$. Another unknown gas at the same temperature and pressure as that of hydrogen is leaked through same hole for $20 \mathrm{~min}$. After the effusion of the gases the mixture exerts a pressure of $6 \mathrm{~atm}$. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is $3 \mathrm{~L}$. What is the molecular weight of the unknown gas?
(1992, 3M)
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Answer:
Correct Answer: 82. $\left(6.2 \times 10^{-21} \mathrm{~J} /\right.$ molecule $)$
Solution:
- Total moles of gas in final mixture $=\frac{p V}{R T}=\frac{6 \times 3}{0.082 \times 300}=0.731$
$\because$ Mole of $\mathrm{H}_{2}$ in the mixture $=0.70$
$\therefore \quad$ Mole of unknown gas $(X)=0.031$
Because both gases have been diffused for same time
$$ \begin{array}{rlrl} & & \frac{r\left(\mathrm{H}_{2}\right)}{r(X)} & =\frac{0.70}{0.031}=\sqrt{\frac{M}{2}} \ \Rightarrow & M & =1020 \mathrm{~g} \mathrm{~mol}^{-1} \end{array} $$
