The rate of diffusion of methane at a given temperature is twice that of a gas $^{'} X^{'}$ . The molecular weight of gas $^{'} X^{'}$ is:

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Solution

The correct option is A 64
According to Graham's law of diffusion,
$r a t e \propto \frac{1}{\sqrt{M o l e c u l a r w e i g h t}}$
$\frac{{r a t e}_{1}}{{r a t e}_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
${(r a t e)}_{m e t h a n e} = 2 \times {(r a t e)}_{a n o t h e r g a s (X)} {(r a t e)}_{g a s X} = x \frac{{(r a t e)}_{m e t h a n e}}{{(r a t e)}_{a n o t h e r g a s (X)}} = \sqrt{\frac{M o l e c u l a r w e i g h t o f X}{16}} ⟹ \frac{2}{1} = \sqrt{\frac{M o l e c u l a r w e i g h t o f X}{16}}$
$⟹$ Molecular weight of gas $X = 64$ g/mole

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