Sulphur dioxide (SO_2) and oxygen (O_2) were | 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 $(M)$ of the gas

Vedclass · Accepted Answer

$14.1$

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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 $(M)$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.The rate of diffusion is defined as $r = \frac{V}{t}$,where $V$ is the volume and $t$ is the time.Thus,$\frac{r_{SO_2}}{r_{O_2}} = \frac{V_{SO_2} / t_{SO_2}}{V_{O_2} / t_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{SO_2}}}$.Given: $V_{SO_2} = 20 \ dm^3$,$t_{SO_2} = 60 \ s$,$t_{O_2} = 30 \ s$.Molar mass of $SO_2 = 32 + (2 	imes 16) = 64 \ g/mol$.Molar mass of $O_2 = 2 	imes 16 = 32 \ g/mol$.Substituting the values: $\frac{20 / 60}{V_{O_2} / 30} = \sqrt{\frac{32}{64}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.$\frac{1/3}{V_{O_2}/30} = \frac{1}{\sqrt{2}} \implies \frac{10}{V_{O_2}} = \frac{1}{1.414}$.$V_{O_2} = 10 	imes 1.414 = 14.14 \ dm^3 \approx 14.1 \ dm^3$.

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