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The rate of diffusion of an unknown gas at $5$ bar is half of the rate of diffusion of $N_{2} O$ gas at $4$ bar. All other conditions are identical. The molar mass of unknown gas (in $gm / mol$ ) is :

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$275$

$273$

$274$

$276$

answer is C.

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Detailed Solution

$r_{gas} = \frac{1}{2} r_{N_{2} O} \frac{r_{gas}}{r_{N_{2} O}} = \frac{1}{2} = \frac{5}{4} \sqrt{\frac{44}{M_{gas}}} \frac{2}{5} = \sqrt{\frac{44}{M_{gas}}} M_{gas} = 275$

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