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Q.

If molar ratio of $C H_{4} (g)$ and $H e (g)$ in the shown container is 1 : 3 then the composition (mole ratio of He and $C H_{4}$ ) of gaseous mixture effusing out initially when valve is opened, is x : 1. The value of x is

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answer is 6.

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Detailed Solution

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$\frac{n_{C H_{4}} / t}{n_{H e} / t} = \frac{P_{C H_{4}}}{P_{H e}} \sqrt{\frac{M_{H e}}{M_{C H_{4}}}}$

$= \frac{1}{3} \times \sqrt{\frac{4}{16}} = \frac{1}{6}$

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