Question

The rates of diffusion of two gases A and B are in the ratio $1:4$. If the ratio of their masses present in the mixture is $2:3$. FInd the ratio of their mole fraction. $(9^{1/3}=2.08)$.

Answer 1

Here let mass of mixture=m

mass of A= $\frac{2m}{5}$

mass of B= $\frac{3m}{5}$

According to Grahams law of diffusion;

$\frac{RA}{RB}=\sqrt{\frac{MB}{MA}}$

$\frac{1}{4}=\sqrt{\frac{MB}{MA}}$

$\frac{1}{16}=\frac{MB}{MA}$

$MA=16MB$

Here, mole fraction of A= $=\frac{\frac{2m}{5MA}}{(\frac{2m}{5MA}+\frac{3m}{5MB})}$

$=\frac{\frac{2m}{5MA}}{(\frac{2m}{5MA}+\frac{3m}{5\left(\frac{MA}{16}\right)})}$

$=\frac{\frac{2}{5}}{\frac{(\frac{2}{5}+\frac{48}{5})\times 2}{50}}=0.04$

mole fraction of B= 1-mole fraction of $A=1-0.04$ = $0.96$

Hence, the answer is $0.96$