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Q.

A 4 : 1 molar mixture of $H e$ and $C H_{4}$ is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially?

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a

8 : 1

b

4 : 1

c

1 : 4

d

1 : 8

answer is A.

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Detailed Solution

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Applying Graham’s law of effusion, we get
$\frac{r_{H e}}{r_{C H_{4}}} = \sqrt{\frac{M_{C H_{4}}}{M_{H e}}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2$
i.e., He diffuses two times faster than $C H_{4} .$
Therefore, The molar ratio of $H e$ and $C H_{4}$ effusing out initially is 8 : 1.

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