Two gases A and B having the same volume,effu | 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 $M$ and directly pro

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$18$

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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 $M$ and directly proportional to the volume $V$ effused per unit time $t$: $r = \frac{V}{t} \propto \frac{1}{\sqrt{M}}$Since the volumes are the same,we have $\frac{V}{t_A} 	imes \sqrt{M_A} = \frac{V}{t_B} 	imes \sqrt{M_B}$,which simplifies to $\frac{t_B}{t_A} = \sqrt{\frac{M_B}{M_A}}$.Given $t_A = 30 	ext{ min}$,$t_B = 10 	ext{ min}$,and $M_A = 162$.Substituting the values: $\frac{10}{30} = \sqrt{\frac{M_B}{162}}$.$\frac{1}{3} = \sqrt{\frac{M_B}{162}}$.Squaring both sides: $\frac{1}{9} = \frac{M_B}{162}$.$M_B = \frac{162}{9} = 18$.

Two gases $A$ and $B$ having the same volume,effuse through an orifice in $30$ and $10$ minutes respectively. The molar mass of $A$ is $162$. The molar mass of $B$ will be

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