$2 K M n O_{4} - --------- - + 5 H_{2} S O_{4} + 10 F e S O_{4} ⟶ 5 {F e}_{2} {(S O_{4})}_{3} + K_{2} S O_{4} + 2 M n S O_{4} + 5 H_{2} O$
The equivalent weight of the underlined species (by taking its molecular weight as $M$ ) is :

E = \frac{M}{2}

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E = \frac{M}{5}

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E = \frac{M}{3}

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none of the above

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Solution

The correct option is B $E = \frac{M}{5}$
The oxidation number of Mn in $K M n O_{4}$ is $+ 7$ .
The oxidation number of Mn in $M n S O_{4}$ is $+ 2$ .
Thus, the oxidation number of Mn changes by $5$ .
$Equivalent weight = \frac{Molecular weight}{Change in the oxidation number}$
Hence, $E = \frac{M}{5}$

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