40 mL of hydrogen diffuses through a fine ho | vedclass

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(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t}$,where $V$ is the volume and $t$ is the time.Also,$r

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$50$

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(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is given by $r = \frac{V}{t}$,where $V$ is the volume and $t$ is the time.Also,$r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass.Therefore,$\frac{V_1 / t_1}{V_2 / t_2} = \sqrt{\frac{M_2}{M_1}}$.Given: $V_{H_2} = 40 \ mL$,$t_{H_2} = 25 \ min$,$M_{H_2} = 2 \ g/mol$.$V_{O_2} = 20 \ mL$,$M_{O_2} = 32 \ g/mol$.Substituting the values: $\frac{40 / 25}{20 / t_{O_2}} = \sqrt{\frac{32}{2}}$.$\frac{40}{25} 	imes \frac{t_{O_2}}{20} = \sqrt{16} = 4$.$1.6 	imes \frac{t_{O_2}}{20} = 4$.$t_{O_2} = \frac{4 	imes 20}{1.6} = \frac{80}{1.6} = 50 \ min$.

$40 \ mL$ of hydrogen diffuses through a fine hole in $25 \ min$. How much time (in $min$) will $20 \ mL$ of oxygen take to diffuse under the same conditions?

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