Question

The time taken for a certain volume of gas to diffuse through a small hole was $2\text{min}$. Under similar conditions an equal volume of oxygen took $4\text{min}$ to pass. The molecular mass of the gas is:

• A. $32.0\text{g}$
• B. $16.0\text{g}$
• C. $8.0\text{g}$
• D. $4.0\text{g}$

Answer 1

The correct option is C $8.0\text{g}$

Graham's law of effusion,
$r\propto \frac{1}{\sqrt{M}}$
Where $r=$ rate of effusion
$M=$ molar mass of gas
Also,
$r=\frac{\text{Volume of gas effused}\left(V\right)}{\text{Time taken}\left(t\right)}$
$\therefore r=\frac{V}{t}$
$\therefore \frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}$
$\Rightarrow \frac{\frac{V_A}{t_A}}{\frac{V_B}{t_B}}=\sqrt{\frac{M_B}{M_A}}$
Where $V_A$ and $V_B$ are the volume of gas $A$ and gas $B$.
$t_A$ and $t_B$ are time taken by gas for effusion.
According to question,
$V_A=V_B=V$
$t_A=2\text{min}$
$t_B=4\text{min}$
$M_B=32\text{g}$ (Molar mass of $O_2$)
$\therefore \frac{\frac{V}{2}}{\frac{V}{4}}=\sqrt{\frac{32}{M_A}}$
$\Rightarrow \frac{4}{2}=\sqrt{\frac{32}{M_A}}$
$\Rightarrow M_A=8\text{g}$