The equivalent mass of potassium permanganate in alkaline medium is:

\frac{molar mass}{2}

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\frac{molar mass}{3}

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\frac{molar mass}{4}

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Molar mass itself

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Solution

The correct option is A Molar mass itself
In alkaline medium,
$+ 7 2 K M n O_{4} + 2 K O H \to + 6 2 K_{3} M n O_{4} + H_{2} O + [O]$
Change in oxidation state of $M n = (+ 7) - (+ 6)$ $= + 1$
$∴$ Equivalent weight of $K M n O_{4}$
$= \frac{molar mass}{change in oxidation state}$
$= \frac{molar mass}{1} =$ molar mass.

Hence, the correct option is $D$ .

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Similar questions

F e S_{2} + K M n O_{4} + H^{+} \to F e^{+ 3} + S O_{2} + M n^{+ 2} + H_{2} O

F e S_{2}

{C r_{2} O_{7}}^{2 -} \to C r^{3 +}

{C r_{2} O_{7}}^{2 -} = \frac{Molar Mass}{3}

2 N a O H + H_{3} P O_{4} \to N a_{2} H P O_{4} + 2 H_{2} O

H_{3} P O_{4}

\frac{M}{3}

= \frac{Molar mass}{n-factor}

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