The rate of diffusion of H_2 is ....... | vedclass

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(B) 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 \propt

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$1.4$ times that of $He$

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(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.For $H_2$,$M_1 = 2 \ g/mol$. For $He$,$M_2 = 4 \ g/mol$.$\frac{r_{H_2}}{r_{He}} = \sqrt{\frac{M_{He}}{M_{H_2}}} = \sqrt{\frac{4}{2}} = \sqrt{2} \approx 1.414$.Therefore,the rate of diffusion of $H_2$ is $1.4$ times that of $He$.

The rate of diffusion of $H_2$ is .......

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