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Graham's Law of Effusion

The rate of d...

Question

The rate of diffusion of methane at a given temperature is twice that of gas X. The molar mass of X is?

A

64

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B

32

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C

4

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D

8

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Solution

The correct option is A
64

$\frac{R_{C H_{4}}}{R_{X}} = \sqrt{\frac{M_{X}}{M_{C H_{4}}}} 2 = \sqrt{\frac{M}{16}} M = 64 g / m o l$

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Graham's Law of Effusion

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