Two gases X and Y are taken in equal molar pr | vedclass

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(A) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$.Given $\frac{r_X}{r_Y} = \frac{1}{2}$.Therefore,$\frac

Vedclass · Accepted Answer

$1/7$

Vedclass · Answer

(A) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$.Given $\frac{r_X}{r_Y} = \frac{1}{2}$.Therefore,$\frac{r_X}{r_Y} = \sqrt{\frac{M_Y}{M_X}} = \frac{1}{2}$.Squaring both sides,$\frac{M_Y}{M_X} = \frac{1}{4}$,which implies $M_X = 4M_Y$.Let $M_Y = M$,then $M_X = 4M$.Moles of $X$ $(n_X)$ = $\frac{2}{4M} = \frac{0.5}{M}$.Moles of $Y$ $(n_Y)$ = $\frac{3}{M}$.Mole fraction of $X$ $(x_X)$ = $\frac{n_X}{n_X + n_Y} = \frac{0.5/M}{0.5/M + 3/M} = \frac{0.5}{3.5} = \frac{5}{35} = \frac{1}{7}$.

Two gases $X$ and $Y$ are taken in equal molar proportions. The ratio of their rates of diffusion is $1 : 2$. What will be the mole fraction of $X$ in a mixture containing $2 \ g$ of $X$ and $3 \ g$ of $Y$?

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