1 L each of gases A and B diffused through a | 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 $(M)$: $r \prop

Vedclass · Accepted Answer

$1:4$

Vedclass · Answer

(A) According to Graham's Law of diffusion, the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$: $r \propto \frac{1}{\sqrt{M}}$.Since the volume $(V)$ is the same for both gases, the rate $r = \frac{V}{t}$.Therefore, $\frac{r_A}{r_B} = \frac{t_B}{t_A} = \sqrt{\frac{M_B}{M_A}}$.Given $t_A = 15 \ 	ext{min}$ and $t_B = 30 \ 	ext{min}$, we have $\frac{30}{15} = \sqrt{\frac{M_B}{M_A}}$.$2 = \sqrt{\frac{M_B}{M_A}}$.Squaring both sides, $4 = \frac{M_B}{M_A}$, which implies $\frac{M_A}{M_B} = \frac{1}{4}$.Thus, the ratio of molecular weight of $A$ and $B$ is $1:4$.

$1 \ L$ each of gases $A$ and $B$ diffused through a membrane in $15$ and $30$ minutes, respectively, under identical conditions. What is the ratio of molecular weight of $A$ and $B$?

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