The ratio of rates of diffusion of gases A an | vedclass

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(C) According to Graham's Law of diffusion,the rate of diffusion $(R)$ is inversely proportional to the square root of the molecular weight $(M)$: $R

Vedclass · Accepted Answer

$16$

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(C) According to Graham's Law of diffusion,the rate of diffusion $(R)$ is inversely proportional to the square root of the molecular weight $(M)$: $R \propto \frac{1}{\sqrt{M}}$.Therefore,the ratio of the rates of diffusion for gases $A$ and $B$ is given by: $\frac{R_A}{R_B} = \sqrt{\frac{M_B}{M_A}}$.Given $\frac{R_A}{R_B} = \frac{1}{0.707}$ and $M_B = 32$,we substitute these values into the equation:$\frac{1}{0.707} = \sqrt{\frac{32}{M_A}}$.Squaring both sides,we get: $(\frac{1}{0.707})^2 = \frac{32}{M_A}$.Since $0.707 \approx \frac{1}{\sqrt{2}}$,then $(\frac{1}{0.707})^2 \approx 2$.So,$2 = \frac{32}{M_A}$,which gives $M_A = \frac{32}{2} = 16$.

The ratio of rates of diffusion of gases $A$ and $B$ is $1 : 0.707$. If the molecular weight of $B$ is $32$,the molecular weight of $A$ is

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