Question

The gas which diffuses twice as quickly as $S{O}_2$ is:

• A. $C{H}_4$
• B. $H_2$
• C. $O_2$
• D. $He$

Answer 1

Answer: (A)

$C{H}_4$

According to Graham's law, the rate of diffusion is inversely proportional to the square root of molecular weight. Hence, methane diffuses twice as fast as sulphur dioxide.

Let $x$ be the gas which diffuses twice as quickly as $S{O}_2.$
$r\propto \frac{1}{\sqrt{M}};$ $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$
$⟹\frac{r_{S{O}_2}}{r_x}=\sqrt{\frac{M_x}{M_{S{O}_2}}}$
$⟹\frac{1}{2}=\sqrt{\frac{M_x}{64}};$ $\frac{1}{4}=\frac{M_x}{64}$
$⟹M_x=\frac{64}{4}=16$
$\therefore$ $x$ is $C{H}_4$ as its molar mass is 16.