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(B) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M_W$: $\frac{r_1}{r

Vedclass · Accepted Answer

$36$

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(B) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M_W$: $\frac{r_1}{r_2} = \sqrt{\frac{M_{W_2}}{M_{W_1}}}$.Since rate $r = \frac{V}{t}$,for the same volume $V$,the expression becomes $\frac{t_2}{t_1} = \sqrt{\frac{M_{W_1}}{M_{W_2}}}$.Given that the time taken by the gas $(t_1)$ is $3$ times the time taken by helium $(t_2)$,we have $\frac{t_1}{t_2} = 3$.Substituting the values: $3 = \sqrt{\frac{M_{W_1}}{4}}$.Squaring both sides: $9 = \frac{M_{W_1}}{4}$.Therefore,$M_{W_1} = 9 	imes 4 = 36 \ u$.

$A$ certain gas takes three times as long to effuse out as helium. Its molecular mass will be ........... $u$.

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