The rate of diffusion of a gas (molar mass = | vedclass

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(D) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r_1 / r_

Vedclass · Accepted Answer

$0.5$

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(D) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r_1 / r_2 = \sqrt{M_2 / M_1}$.Given: Molar mass of gas $(M_1 = 128)$,Molar mass of oxygen $(M_2 = 32)$.Substituting the values: $r_{gas} / r_{O_2} = \sqrt{32 / 128} = \sqrt{1 / 4} = 0.5$.Therefore,the rate of diffusion of the gas is $0.5$ times that of oxygen.

The rate of diffusion of a gas (molar mass = $128$) compared to oxygen is:

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