The number of moles of $K M n O_{4}$ reduced by one mole of KI in alkaline medium is

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\frac{1}{5}

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Solution

The correct option is A 1
+7 +6
$K M n O_{4} O H^{-} - -- \to K_{2} M n O_{4}$
Change in oxidation number of Mn in basic medium is 1.
$∴$ Moles of KI is equal to moles of $K M n O_{4}$

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