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In neutral or faintly alkaline medium, $\mathrm{MnO}_4^{-}$oxidizes $\mathrm{I}^{-}$to iodate. What is the volume (in $\mathrm{L}$ ) of $0.02 \mathrm{M} \mathrm{KMnO}_4$ required to completely convert $1 \mathrm{~L}$ of $0.5 \mathrm{M} \mathrm{KI}$ solution to iodate in neutral or faintly alkaline medium?
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The correct answer is:
50
$2 \mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{KI} \rightarrow 2 \mathrm{MnO}_2+2 \mathrm{KOH}+\mathrm{KIO}_3$
$\Rightarrow 1$ mole of $\mathrm{I}^{-}$requires 2 moles of $\mathrm{MnO}_4^{-}$.
Now, number of moles of $1 \mathrm{~L} \times 0.5 \mathrm{M}=0.5 \mathrm{~mol}$.
$\Rightarrow$ Number of moles of $\mathrm{MnO}_4^{-}$required
$=2 \times 0.5=1.0 \mathrm{~mol}$.
$\Rightarrow$ Volume $=\frac{\mathrm{n}}{\mathrm{M}}=\frac{1.0}{0.02}=50 \mathrm{~L}$
$\Rightarrow 1$ mole of $\mathrm{I}^{-}$requires 2 moles of $\mathrm{MnO}_4^{-}$.
Now, number of moles of $1 \mathrm{~L} \times 0.5 \mathrm{M}=0.5 \mathrm{~mol}$.
$\Rightarrow$ Number of moles of $\mathrm{MnO}_4^{-}$required
$=2 \times 0.5=1.0 \mathrm{~mol}$.
$\Rightarrow$ Volume $=\frac{\mathrm{n}}{\mathrm{M}}=\frac{1.0}{0.02}=50 \mathrm{~L}$
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