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. The ratio of their mole
fraction is$\left(9^{t/3}=2.08\right)$.

Answer 1

Assume the mass of mixture= $m$

Mass of A= $2m/5$

Mass of B= $3m/5$

$R_A/R_B=\sqrt{M_B/M_A}$ (Graham's law of diffusion)

$\Rightarrow \frac{1}{4}=\sqrt{\frac{M_B}{M_A}}\Rightarrow M_A=16M_B$

Now, mole fraction of A=$\frac{\frac{2m}{5M_A}}{\frac{2m}{5M_A}+\frac{3m}{5M_B}}$

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

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