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(A) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass.Let $M_{N_2} = M$. Then th

Vedclass · Accepted Answer

$79.19$

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(A) According to Graham's Law of diffusion,the rate of diffusion $r \propto \frac{1}{\sqrt{M}}$,where $M$ is the molar mass.Let $M_{N_2} = M$. Then the molar mass of the other gas is $M_x = 2M$.Given: $r_x = 56 \ mL \ s^{-1}$.Using the formula: $\frac{r_{N_2}}{r_x} = \sqrt{\frac{M_x}{M_{N_2}}}$.$\frac{r_{N_2}}{56} = \sqrt{\frac{2M}{M}} = \sqrt{2}$.$r_{N_2} = 56 	imes \sqrt{2} = 56 	imes 1.414 = 79.184 \approx 79.19 \ mL \ s^{-1}$.

If the rate of diffusion of a gas having twice the molecular mass of nitrogen is $56 \ mL \ s^{-1}$,then the rate of diffusion of nitrogen will be ........... $mL \ s^{-1}$.

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