The rate of diffusion of SO_2 and O_2 are in | vedclass

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(A) 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 \f

Vedclass · Accepted Answer

$1 : \sqrt{2}$

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(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.Therefore,the ratio of the rates of diffusion of $SO_2$ and $O_2$ is given by: $\frac{r_{SO_2}}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{SO_2}}}$.The molar mass of $SO_2$ is $64 \ g/mol$ and the molar mass of $O_2$ is $32 \ g/mol$.Substituting these values: $\frac{r_{SO_2}}{r_{O_2}} = \sqrt{\frac{32}{64}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Thus,the ratio is $1 : \sqrt{2}$.

The rate of diffusion of $SO_2$ and $O_2$ are in the ratio:

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