The ratio of the rate of diffusion of SO_2 an | 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 $(M)$ of the gas

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 $(M)$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.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 $32 + (2 	imes 16) = 64 \ g/mol$.The molar mass of $O_2$ is $2 	imes 16 = 32 \ g/mol$.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 ratio of the rate of diffusion of $SO_2$ and $O_2$ is .......

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