A specific volume of H_2 requires 24 s to di | vedclass

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Question

(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

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(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: $r \propto \frac{1}{\sqrt{M}}$.For two gases $O_2$ and $H_2$,the ratio of their rates of diffusion is given by: $\frac{r_{O_2}}{r_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}$.Given $M_{H_2} = 2 \ g/mol$ and $M_{O_2} = 32 \ g/mol$,we have: $\frac{r_{O_2}}{r_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.Since rate $r = \frac{V}{t}$,for equal volumes $V$,the ratio of rates is inversely proportional to the ratio of times: $\frac{r_{O_2}}{r_{H_2}} = \frac{t_{H_2}}{t_{O_2}}$.Substituting the values: $\frac{1}{4} = \frac{24 \ s}{t_{O_2}}$.Therefore,$t_{O_2} = 24 	imes 4 = 96 \ s$.

$A$ specific volume of $H_2$ requires $24 \ s$ to diffuse out of a container. The time required by an equal volume of $O_2$ to diffuse out under identical conditions is $..... \ s$.

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