Question

A gas flushes out of a container through a pin hole four times quickly as methane. The molecular weight of the gas is

• A. $2\text{g}$
• B. $1\text{g}$
• C. $16\text{g}$
• D. $8\text{g}$

Answer 1

The correct option is B $1\text{g}$

From Graham's law of effusion,
Rate of effusion of gas ; $r\propto \frac{1}{\sqrt{M}}$
where $M=$ Molar mass
Let the molecular weight of unknown gas be $M$
Molecular mass of methane $\left(C{H}_4\right)=16\text{g}$
Now,
$r\propto \frac{1}{\sqrt{M}}$
and $r_{C{H}_4}\propto \frac{1}{\sqrt{M_{C{H}_4}}}$
$\therefore \frac{r}{r_{C{H}_4}}=\sqrt{\frac{M_{C{H}_4}}{M}}=4$
where $r$ is rate of effusion of unknown gas.
$\Rightarrow \sqrt{\frac{M_{C{H}_4}}{M}}=4$
$\Rightarrow \frac{M_{C{H}_4}}{M}=16$
$\Rightarrow M=\frac{16}{16}=1\text{g}$