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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 $M$: $\frac{r_{H_2

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$16$

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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 $M$: $\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.Given $M_{O_2} = 32 \ g/mol$ and $M_{H_2} = 2 \ g/mol$,we have $\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.Since rate $r = \frac{	ext{mass}}{t}$,for the same time $t$,$\frac{	ext{mass}_{H_2}}{	ext{mass}_{O_2}} = 4$.Given $	ext{mass}_{O_2} = 4 \ g$,then $	ext{mass}_{H_2} = 4 	imes 4 = 16 \ g$.

If $4 \ g$ of oxygen diffuse through a very narrow hole,how much hydrogen would have diffused under identical conditions? (in $g$)

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