If some moles of O_2 diffuse in 18 sec and t | vedclass

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Question

(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ and inversely

Vedclass · Accepted Answer

$\frac{45^2}{18^2} 	imes 32$

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(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ and inversely proportional to the time $t$ taken for the same amount of gas to diffuse: $r \propto \frac{1}{\sqrt{M}}$ and $r \propto \frac{1}{t}$.Therefore,$\frac{r_1}{r_2} = \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}$.Given $t_{O_2} = 18 \ sec$,$M_{O_2} = 32 \ g/mol$,$t_g = 45 \ sec$,and $M_g = ?$.Substituting the values: $\frac{45}{18} = \sqrt{\frac{M_g}{32}}$.Squaring both sides: $(\frac{45}{18})^2 = \frac{M_g}{32}$.Thus,$M_g = 32 	imes \frac{45^2}{18^2}$.

If some moles of $O_2$ diffuse in $18 \ sec$ and the same moles of another gas diffuse in $45 \ sec$,what is the molecular weight of the unknown gas?

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