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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 volume and $t$ is time.Graham's Law st

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$64$

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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 volume and $t$ is time.Graham's Law states: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$,where $M$ is the molar mass.Substituting $r = \frac{V}{t}$,we get: $\frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}}$.Given: $V_1 = 50 \ mL$,$t_1 = 20 \ min$,$M_1 (H_2) = 2 \ g/mol$,$V_2 = 40 \ mL$,$M_2 (O_2) = 32 \ g/mol$.$\frac{50/20}{40/t_2} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.$\frac{50}{20} 	imes \frac{t_2}{40} = 4$.$2.5 	imes \frac{t_2}{40} = 4$.$t_2 = \frac{4 	imes 40}{2.5} = \frac{160}{2.5} = 64 \ min$.

$50 \ mL$ of hydrogen gas diffuses through a small hole in a container in $20 \ min$. How much time (in $min$) will $40 \ mL$ of oxygen take to diffuse under the same conditions?

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