At STP,it takes 1200 s for 0.48 g of O_2 to | vedclass

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(A) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r_1 / r_2 = \s

Vedclass · Accepted Answer

$286.5$

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(A) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r_1 / r_2 = \sqrt{M_2 / M_1}$.Since the time $t$ is the same for both gases,the rate is proportional to the amount of gas effused: $n_1 / n_2 = \sqrt{M_2 / M_1}$.Given: $n(O_2) = 0.48 \ g / 32 \ g/mol = 0.015 \ mol$.Molar mass of $O_2 = 32 \ g/mol$,Molar mass of $CO_2 = 44 \ g/mol$.$0.015 / n(CO_2) = \sqrt{44 / 32} = \sqrt{1.375} \approx 1.1726$.$n(CO_2) = 0.015 / 1.1726 \approx 0.01279 \ mol$.At $STP$,$1 \ mol$ of gas occupies $22400 \ mL$.Volume of $CO_2 = 0.01279 \ mol 	imes 22400 \ mL/mol \approx 286.5 \ mL$.

At $STP$,it takes $1200 \ s$ for $0.48 \ g$ of $O_2$ to effuse through a pinhole. How many $mL$ of $CO_2$ gas will effuse under the same conditions and in the same time?

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