The rate of diffusion of a gas having a molar | vedclass

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Question

(D) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of its molar mass $(M)$: $r \propt

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(D) According to Graham's Law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of its molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.For two gases,the ratio of their rates of diffusion is given by: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Here,$M_1 = 128 \ g/mol$ (gas) and $M_2 = 32 \ g/mol$ (oxygen).Substituting the values: $\frac{r_{gas}}{r_{O_2}} = \sqrt{\frac{32}{128}} = \sqrt{\frac{1}{4}} = 0.5$.Thus,the rate of diffusion of the gas is $0.5$ times that of oxygen.

The rate of diffusion of a gas having a molar mass of $128 \ g/mol$ with respect to oxygen is ........... times.

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