Q22 / 160 · TS EAMCET 2014 · Chemistry
$\mathrm{KMnO}_4$ reacts with $\mathrm{KI}$, in basic medium to form $\mathrm{I}_2$ and $\mathrm{MnO}_2$. When $250 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{KI}$ solution is mixed with $250 \mathrm{~mL}$ of $0.02 \mathrm{M}$ $\mathrm{KMnO}_4$ in basic medium, what is the number of moles of $\mathrm{I}_2$ formed?
- A$0.015$
- Correct answerB$0.0075$
- C$0.005$
- D$0.01$
Solution
$\begin{array}{r}\lceil\text {-factor }=1 \longrightarrow \\ \mathrm{MnO}_4^{-}+\mathrm{I}^{-} \rightleftharpoons \mathrm{MnO}_2+\mathrm{I}_2 \\ \bigsqcup_{n \text {-factor }=3}^{\rightleftharpoons}\end{array}$
Number of milli equivalents of
$\mathrm{MnO}_4^{-}=0.02 \times 3 \times 250=15$
Number of milli equivalents of
$\mathrm{I}_2=0.1 \times 1 \times 250=25$
Thus, here $\mathrm{MnO}_4^{-}$is limiting reagent.
$\therefore$ Number of milli equivalents of $\mathrm{I}_2$ formed $=$
Number of milli equivalent of $\mathrm{MnO}_4^{-}=15$
or number of equivalent of $\mathrm{I}_2$ formed
$=\frac{15}{1000}=0.015$
$\begin{aligned}
\therefore \text { Number of moles of } \mathrm{I}_2 \text { formed } & =\frac{0.015}{2} \\
& =0.0075
\end{aligned}$
Hence, the correct answer is option: 'B'.