Question

The ratio of velocities of the 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

$\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}=\frac{1}{4}$
$\text{or,}\frac{M_B}{M_A}=\frac{1}{16}$
Let masses $W_A$ of $A$ and $W_B$ of $B$ be present respectively.
$\text{So,}\frac{W_A}{W_B}=\frac{2}{3}$ (given)
$\text{Mole fraction of}A,{\chi }_A=\frac{\frac{W_A}{M_A}}{\frac{W_A}{M_A}+\frac{W_B}{M_B}}$
$\text{Mole fraction of}B,{\chi }_B=\frac{\frac{W_B}{M_B}}{\frac{W_A}{M_A}+\frac{W_B}{M_B}}$
$\frac{{\chi }_A}{{\chi }_B}=\frac{W_A}{M_A}\times \frac{M_B}{W_B}$
$\text{or}\frac{W_A}{W_B}\times \frac{M_B}{M_A}=\frac{2}{3}\times \frac{1}{16}=\frac{1}{24}$