At a definite temperature, rate of diffusion of $C H_{4}$ gas is twice that of gas X, then calculate molecular mass of X?

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Solution

The correct option is A 64
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{d_{2}}{d_{2}}}$ (by Graham's law of diffusion)
where, r is rate of diffusion
and d is density
and $d \propto M (d = \frac{M}{V})$
So,
$\frac{r_{1}}{r_{2}} = \sqrt{\frac{d_{2}}{d_{1}}} = \sqrt{\frac{M_{2}}{M_{1}}}$
Given, rate of diffusion of methane at a given temperature is twice that of X
$r_{C H_{2}} = 2 r_{X} \frac{r_{C H_{2}}}{r_{X}} = \sqrt{\frac{M_{X}}{M_{C H_{2}}}} \frac{2 r_{X}}{r_{X}} = \sqrt{\frac{M_{X}}{16}} 4 = \frac{M_{X}}{16} 16 \times 4 = M_{X}$
$M_{X} = 64 g m o l^{- 1}$ .

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