Question

A mixture of $S{O}_2$, and $O_2$ in the molar ratio $16:1$ is diffused through a pin hole for successive effusions three times to give a molar ratio $1:1$ of diffused mixture. Which one are not correct if diffusion is made at same $P$ and $T$ in each operation?
$I.$ Eight operation are needed to get $1:1$ molar ratio.
$II.$ Rate of diffusion for $S{O}_2:O_2$ after eight operations in $\mathrm{0.707.}$
$III.$ Six operations are needed to get $2:1$ molar ratio for $S{O}_2$ and $O_2$ in diffusion mixture.
$IV.$ Rate of diffusion for $S{O}_2$ and $O_2$ after six operations is $\mathrm{2.41.}$

• A. $I,II,III$
• B. $II,III$
• C. $I,III$
• D. $IV$

Answer 1

The correct option is A $I,II,III$

$(f_1{)}^X=\frac{n_{S{O}_2}^{\prime }}{n_{O_2}^{\prime }}\times \frac{n_{O_2}}{n_{S{O}_2}}$, where $n_{S{O}_2}$ and $n_{O_2}$ are moles present initially.
or $X\log f_1=\log [\frac{n_{S{O}_2}^{\prime }}{n_{O_2}^{\prime }}\times \frac{n_{O_2}}{n_{S{O}_2}}]$

$\therefore X\log \sqrt{\frac{M_{O_2}}{M_{S{O}_2}}}=\log [\frac{n_{S{O}_2}^{\prime }}{n_{O_2}^{\prime }}\times \frac{n_{O_2}}{n_{S{O}_2}}]$
$X\log \sqrt{\frac{32}{64}}=\log \frac{1}{1}\times \frac{1}{16}$

$\therefore X=8;also\frac{n_1}{n_2}=\frac{r_1}{r_2}=0.707$
If $X=6,then6\log \sqrt{\frac{32}{64}}=\log [\frac{n_{S{O}_2}^{\prime }}{n_{O_2}^{\prime }}\times \frac{n_{O_2}}{n_{S{O}_2}}]$
$=\log [\frac{n_{S{O}_2}^{\prime }}{n_{O_2}^{\prime }}\times \frac{1}{16}]$
$=\frac{n_{S{O}_2}^{\prime }}{n_{O_2}^{\prime }}=2:1$
Rate of diffusion is $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$, i.e., 0.707 in each operation.