If 4 g of oxygen diffuses through a fine hol | 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 $M$ of the gas: $r

Vedclass · Accepted Answer

$1$

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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 $M$ of the gas: $r \propto \frac{1}{\sqrt{M}}$.Since the time $t$ is the same for both gases,the amount diffused $n$ (in moles) is proportional to the rate of diffusion: $n \propto r$.Therefore,$\frac{n_{H_2}}{n_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.The molar mass of $O_2$ is $32 \ g/mol$ and $H_2$ is $2 \ g/mol$.Let $w$ be the mass of $H_2$ in grams. The number of moles $n = \frac{w}{M}$.$\frac{w_{H_2} / 2}{4 / 32} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.$\frac{w_{H_2}}{2} = 4 	imes \frac{4}{32} = 4 	imes 0.125 = 0.5$.$w_{H_2} = 0.5 	imes 2 = 1 \ g$.

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

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