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

One fifth

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five

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One

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Two

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Solution

The correct option is D
Two

In alkaline medium
$2 K M n O_{4} + K I + H_{2} O ⟶ 2 M n O_{2} + 2 K O H + K I O_{3}$

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

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