12 cm^3 of SO_{2(g)} diffused through a poro | vedclass

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Question

(D) According to Graham's law of diffusion, the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$, i.e.,$r

Vedclass · Accepted Answer

$16$

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(D) According to Graham's law of diffusion, the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$, i.e.,$r \propto \frac{1}{\sqrt{M}}$.Also, the rate $(r)$ is directly proportional to the volume $(V)$ diffused per unit time $(t)$, i.e.,$r = \frac{V}{t}$.Therefore, $\frac{r_1}{r_2} = \frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}}$.Given:$V_{SO_2} = 12 \ cm^3$, $t_{SO_2} = 1 \ min$, $M_{SO_2} = 64 \ g \ mol^{-1}$.$V_{gas} = 120 \ cm^3$, $t_{gas} = 5 \ min$, $M_{gas} = ?$.Substituting the values:$\frac{12/1}{120/5} = \sqrt{\frac{M_{gas}}{64}}$.$\frac{12}{24} = \sqrt{\frac{M_{gas}}{64}}$.$0.5 = \sqrt{\frac{M_{gas}}{64}}$.Squaring both sides:$0.25 = \frac{M_{gas}}{64}$.$M_{gas} = 0.25 	imes 64 = 16 \ g \ mol^{-1}$.

$12 \ cm^3$ of $SO_{2(g)}$ diffused through a porous membrane in $1 \ minute$. Under similar conditions $120 \ cm^3$ of another gas diffused in $5 \ minutes$. The molar mass of the gas in $g \ mol^{-1}$ is

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