Question

$0.5$ g mixture of $K_{2}C{r}_2{O}_7$ and $KMn{O}_4$ was treated with excess of $KI$ in acidic medium. Iodine liberated required $150c{m}^3$ of $0.10N$ solution of thiosulphate solution for titration. Find the percentage of $K_{2}C{r}_2{O}_7$ in the mixture.

• A. $14.64$
• B. $34.2$
• C. $65.69$
• D. $50$

Answer 1

Answer: (A)

$14.64$

Reactions of $K_{2}C{r}_2{O}_7$ and $KMn{O}_4$ with $KI$ is given as follows:

$K_{2}C{r}_2{O}_7+7H_{2}S{O}_4+6KI\to 4K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+7H_{2}O+3I_2$
$KMn{O}_4+8H_{2}S{O}_4+10KI\to 6K_{2}S{O}_4+2MnS{O}_4+5I_2$

Thus, equivalent weight of $K_{2}C{r}_2{O}_7=\frac{294}{6}=49$
Equivalent weight of $KMn{O}_4=\frac{158}{5}=31.6$
$\because$ Milli-equivalents of $K_{2}C{r}_2{O}_7$ $+$ Milli-equivalents of $KMn{O}_4$ $=$ Milli-equivalents of $I_2$ $=$ Milli-equivalents of hypo
Let the mass of $K_{2}C{r}_2{O}_7$ be $x$ g.
$\therefore$ Mass of $KMn{O}_4=(0.5-x)$ g
$\frac{x}{49}+\frac{(0.5-x)}{31.6}=150\times 0.1\times {10}^{-3}$ or $x=0.0732$
$\therefore \%$ of $K_{2}C{r}_2{O}_7=\frac{0.0732}{0.5}\times 100=14.64$