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A gaseous mixture containing $\mathrm{He}, \mathrm{CH}_{4}$ and $\mathrm{SO}_{2}$ was allowed to effuse through a fine hole then find what molar ratio of gases coming out initially? (Given mixture contains $\mathrm{He}, \mathrm{CH}_{4}$ and $\mathrm{SO}_{2}$ in $1: 2: 3$ mole ratio).
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The correct answer is:
$4: 4: 3$
$\begin{array}{l}
\frac{\mathrm{n}_{\mathrm{He}}^{\prime}}{\mathrm{n}_{\mathrm{CH}_{4}}}=\frac{1}{2} \sqrt{\frac{16}{4}}=\frac{1}{1} \\
\frac{\mathrm{n}_{\mathrm{He}}^{\prime}}{\mathrm{n}^{\prime} \mathrm{SO}_{2}}=\frac{1}{3} \sqrt{\frac{64}{4}}=\frac{4}{3}
\end{array}$
So, molar ratio will be,
$\mathrm{n}^{\prime} \mathrm{He}: \mathrm{n}^{\prime} \mathrm{CH}_{4}: \mathrm{n}^{\prime} \mathrm{SO}_{2}=4: 4: 3 \text {. }$
\frac{\mathrm{n}_{\mathrm{He}}^{\prime}}{\mathrm{n}_{\mathrm{CH}_{4}}}=\frac{1}{2} \sqrt{\frac{16}{4}}=\frac{1}{1} \\
\frac{\mathrm{n}_{\mathrm{He}}^{\prime}}{\mathrm{n}^{\prime} \mathrm{SO}_{2}}=\frac{1}{3} \sqrt{\frac{64}{4}}=\frac{4}{3}
\end{array}$
So, molar ratio will be,
$\mathrm{n}^{\prime} \mathrm{He}: \mathrm{n}^{\prime} \mathrm{CH}_{4}: \mathrm{n}^{\prime} \mathrm{SO}_{2}=4: 4: 3 \text {. }$
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