Question

Calculate relative rate of effusion of $S{O}_2$ to $C{H}_4$ under given condition:

( i ) Under similar condition of pressure & temperature
( ii ) Through a container containing $S{O}_2$ and $C{H}_4$ in 3 : 2 mass ratio
( iii ) If the mixture obtained by effusing out a mixture $(n_{S{O}_2}/n_{C{H}_4}=8/1)$ for three effusing steps.

• A. ( i ) $\frac{1}{2}$ ( ii ) $\frac{3}{16}$ ( iii ) $\frac{1}{2}$
• B. ( i ) $\frac{1}{4}$ ( ii ) $\frac{3}{8}$ ( iii ) $\frac{1}{4}$
• C. ( i ) $\frac{1}{6}$ ( ii ) $\frac{5}{16}$ ( iii ) $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is A ( i ) $\frac{1}{2}$ ( ii ) $\frac{3}{16}$ ( iii ) $\frac{1}{2}$

The relative rate of effusion is inversely proportional to the square root of molar mass.It is directly proportional to the number of moles .The molar masses of sulphur dioxide and methane are 64 g/mol and 16 g/mol respectively.
(i) Under similar condition of pressure & temperature
$\frac{r_{S{O}_2}}{r_{C{H}_4}}=\sqrt{\frac{M_{C{H}_4}}{M_{S{O}_2}}}=\sqrt{\frac{16}{64}}=\frac{1}{2}$
( ii ) Through a container containing $S{O}_2$ and $C{H}_4$ in 3 : 2 mass ratio
$\frac{r_{S{O}_2}}{r_{C{H}_4}}=\sqrt{\frac{n_{S{O}_2}{M}_{C{H}_4}}{n_{C{H}_4}{M}_{S{O}_2}}}=\sqrt{\frac{\frac{3}{64}\times 16}{\frac{2}{16}\times 64}}=\frac{3}{16}$
( iii ) If the mixture obtained by effusing out a mixture $(n_{S{O}_2}/n_{C{H}_4}=8/1)$ for three effusing steps.
$\frac{r_{S{O}_2}}{r_{C{H}_4}}=\sqrt{\frac{M_{C{H}_4}}{M_{S{O}_2}}}=\sqrt{\frac{16}{64}}=\frac{1}{2}$