$3 M n O_{2} - ------- - + 6 K O H + K C l O_{3} ⟶ 3 K_{2} M n O_{4} + K C l + 3 H_{2} O$
The equivalent weight of the underlined species by taking its molecular weight as $M$ is :

E = M

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E = \frac{M}{3}

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E = \frac{M}{6}

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E = \frac{M}{2}

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Solution

The correct option is D $E = \frac{M}{2}$
The oxidation number of Mn in $M n O_{2}$ is $+ 4$ .
The oxidation number of Mn in $K_{2} M n O_{4}$ is $+ 6$ .
Thus, the oxidation number of $M n$ changes by $2$ .
$Equivalent weight = \frac{Molecular weight}{Change in the oxidation number}$
Hence, $E = \frac{M}{2}$

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