The rate of diffusion of methane at a given temperature and pressure is twice that of a gas $X$ . The molecular weight of $X$ will be:

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Solution

The correct option is B 64
The rate of diffusion of methane is twice that of a gas $X$ . According to Graham's law of diffusion,
$\frac{{r a t e}_{1}}{{r a t e}_{2}} = \sqrt{\frac{M_{2}}{M_{1}}} ⟹ \frac{2}{1} = \sqrt{\frac{M o l a r m a s s X}{16}}$
$∴$ Molar mass of $X = 64$ g/mol

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