Question

The ratio of velocities of diffusion of gases $A$ and $B$ is $1:4$. If the ratio of their masses present in the mixture is $2:3$, calculate the ratio of their mole fractions.

Answer 1

$\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}=\frac{1}{4}$
or $\frac{M_B}{M_A}=\frac{1}{16}$
Let masses $W_A$ and $W_B$ of $A$ and $B$ be present respectively.
Mole fraction of $A=\frac{\frac{W_A}{M_A}}{\frac{W_A}{M_A}+\frac{W_B}{M_B}}$
Mole fraction of $B=\frac{\frac{W_B}{M_B}}{\frac{W_A}{M_A}+\frac{W_B}{M_B}}$
$Ratio=\frac{W_A}{M_A}\times \frac{M_B}{W_B}$
or $\frac{W_A}{W_B}\times \frac{M_B}{M_A}=\frac{2}{3}\times \frac{1}{16}=\frac{1}{24}$.