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The number of moles of $\mathrm{KMnO}_4$ reduced by one mole of $K I$ in alkaline medium is:
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The correct answer is:
Two
In alkaline medium
$2 \mathrm{KMnO}_4+\mathrm{Kl}+\mathrm{H}_2 \mathrm{O}-\cdots-\mathrm{MnO}_2+2 \mathrm{KOH}+\mathrm{KlO}_3$
$2 \mathrm{KMnO}_4+\mathrm{Kl}+\mathrm{H}_2 \mathrm{O}-\cdots-\mathrm{MnO}_2+2 \mathrm{KOH}+\mathrm{KlO}_3$
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