The rate of diffusion of methane at at given temprature is twice that of gas X. The molecule weight of X is:

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Solution

The correct option is A 64
Let, the rate of diffusion of Methane is $r_{1}$

rate of diffusion of second gas $X$ is $r_{2}$

We know the rate of diffusion is inversely proportional to molecular weight.

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$

$\frac{r_{1}}{2 r_{1}} = \sqrt{\frac{16}{M_{2}}}$

$\frac{1}{4} = \frac{16}{M_{2}}$

$M_{2} = 64$

Hence, the correct option is $A$

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