If 4 g of oxygen diffuses through a narrow h | 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 $M_w$ for a given

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$1$

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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 $M_w$ for a given amount of gas.$\frac{r_{O_2}}{r_{H_2}} = \frac{n_{O_2}}{n_{H_2}} = \sqrt{\frac{M_{w(H_2)}}{M_{w(O_2)}}}$Given: $n_{O_2} = \frac{4 \ g}{32 \ g/mol} = 0.125 \ mol$.Let $w$ be the mass of $H_2$. Then $n_{H_2} = \frac{w}{2 \ g/mol}$.Substituting the values: $\frac{0.125}{w/2} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.$\frac{0.25}{w} = \frac{1}{4} \implies w = 1 \ g$.

If $4 \ g$ of oxygen diffuses through a narrow hole under identical conditions,how many grams of hydrogen will diffuse?

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