Question

$0.32$ g sample of impure $KI$ was dissolved in $1.1$ millimoles of $K_{2}Cr{O}_4$ and $25$ mL of $5$ N $H_{2}S{O}_4$. Iodine formed was expelled off by boiling and the solution is now mixed with excess of pure $KI$ and $I_2$ liberated again is completely reduced by $14$ mL of $0.1NN{a}_2{S}_2{O}_3$. Calculate percentage purity of impure $KI$ sample. The reactions are given below:
$2Cr{O}_4^{2-}+2H^{+}⟶C{r}_2{O}_7^{2-}+H_{2}O$
$C{r}_2{O}_7^{2-}+6I^{-}+14H^{+}⟶2C{r}^{3+}+3I_2+7H_{2}O$

• A. $96$
• B. $91$
• C. $89$
• D. None of these

Answer 1

The correct option is D None of these

$2Cr{O}_4^{2-}+2H^{+}⟶C{r}_2{O}_7^{2-}+H_{2}O$
$\therefore$ millimoles of $C{r}_2{O}_7^{2-}$ formed $=\frac{1}{2}\times$ millimoles of $Cr{O}_4^{2-}$ $=\frac{1}{2}\times 1.1=0.55$
$\therefore$ Meq. of $C{r}_2{O}_7^{2-}$ formed $=0.55\times 6=3.30$ (Valence factor of $C{r}_2{O}_7^{2-}=6$)
Millimoles of $H^{+}$ ($62.5$) are more than millimoles of $Cr{O}_4^{2-}$($1.1$).
Thus, whole of $Cr{O}_4^{2-}$ will be converted to $C{r}_2{O}_7^{2-}$.
Meq. of $C{r}_2{O}_7^{2-}$ left after reaction with impure $KI=$ meq. of $I_2$ formed in $IIstep=$ Meq. of hypo used $=14\times 0.1=1.4$
Meq. of $KI$ in impure sample $=3.30-1.40=1.90$
$\frac{w_{KI}}{E}\times =1.9$
$w_{KI}=\frac{1.9\times 166}{1000}=0.3154$
$\therefore \%$ purity $=\frac{0.3154}{0.32}\times 100=98.56$