2.0 g of H_2 diffuses through a porous conta | vedclass

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(D) 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

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$8$

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(D) 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: $r \propto \frac{1}{\sqrt{M}}$.Since the time $t$ is the same,the rate of diffusion is directly proportional to the mass $w$ diffused: $r = \frac{w}{t}$.Therefore,$\frac{w_1}{w_2} = \sqrt{\frac{M_2}{M_1}}$.Given: $w_{H_2} = 2.0 \ g$,$M_{H_2} = 2 \ g/mol$,$M_{O_2} = 32 \ g/mol$.Substituting the values: $\frac{2.0}{w_{O_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.$w_{O_2} = 2.0 	imes 4 = 8.0 \ g$.

$2.0 \ g$ of $H_2$ diffuses through a porous container in $10 \ minutes$. How many grams of $O_2$ would diffuse from the same container in the same time under similar conditions (in $.0$)?

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