If 4 g of oxygen gas diffuses through a fine | 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 $M$ for a given ti

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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 $M$ for a given time $t$: $\frac{n_1}{n_2} = \sqrt{\frac{M_2}{M_1}}$,where $n$ is the number of moles.Since $n = \frac{w}{M}$,the ratio of weights $w$ is given by: $\frac{w_1}{w_2} = \sqrt{\frac{M_1}{M_2}}$.Here,$w_{O_2} = 4 \ g$,$M_{O_2} = 32 \ g/mol$,and $M_{H_2} = 2 \ g/mol$.Substituting the values: $\frac{4}{w_{H_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.Therefore,$w_{H_2} = \frac{4}{4} = 1 \ g$.

If $4 \ g$ of oxygen gas diffuses through a fine hole,how many grams of hydrogen gas will diffuse under the same conditions?

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