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(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ for the same v

Vedclass · Accepted Answer

$12.25$

Vedclass · Answer

(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$ for the same volume of gas: $\frac{r_A}{r_B} = \sqrt{\frac{M_B}{M_A}}$.Since $r = \frac{V}{t}$ and the volume $V$ is the same,$\frac{r_A}{r_B} = \frac{t_B}{t_A}$.Given $t_A = 20 \ s$,$t_B = 10 \ s$,and $M_A = 49 \ u$,we have:$\frac{10}{20} = \sqrt{\frac{M_B}{49}}$$\frac{1}{2} = \sqrt{\frac{M_B}{49}}$Squaring both sides:$\frac{1}{4} = \frac{M_B}{49}$$M_B = \frac{49}{4} = 12.25 \ u$.

Under the same conditions of pressure and temperature,the same volume of two gases $A$ and $B$ diffuse through a porous partition in $20 \ s$ and $10 \ s$ respectively. The molecular mass of $A$ is $49 \ u$. The molecular mass of $B$ will be ............. $u$.

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