Question

Calculate relative rate of effusion of O${}_2$ to CH${}_4$ through a container containing O${}_2$ and CH${}_4$ in 3 : 2 mass ratio.

• A. $\frac{3\sqrt{2}}{4}$
• B. $\frac{3}{4\sqrt{2}}$
• C. $\frac{3}{2\sqrt{2}}$
• D. None of these

Answer 1

The correct option is B $\frac{3}{4\sqrt{2}}$

The relative rate of effusion are given by the expression $\frac{r_1}{r_2}=\frac{n_1}{n_2}\times \sqrt{\frac{M_2}{M_1}}$.
Here $1$ represents oxygen and $2$ represents methane.

The molecular weight of $O_2$ is $32$ g/mol. The molecular weight of methane is $16$ g/mol.
Oxygen and methane are present in $3:2$ mole ratio.

$\frac{n_1}{n_2}=\frac{\frac{3}{32}}{\frac{2}{16}}$

Substitute values in the above expression.

$\frac{r_1}{r_2}=\frac{\frac{3}{32}}{\frac{2}{16}}\times \sqrt{\frac{16}{32}}=\frac{3}{4\sqrt{2}}$

So, the correct option is $B$