$2 K C l O_{3} - ------- - ⟶ 2 K C l + 3 O_{2}$
The equivalent weight of the underlined species (by taking its molecular weight as $M$ ) is :

E = \frac{M}{6}

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E = \frac{M}{4}

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E = \frac{M}{3}

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none of these

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Solution

The correct option is A $E = \frac{M}{6}$
The oxidation number of Cl in $K C l O_{3}$ is $5$ .
The oxidation number of Cl in $K C l$ is $- 1$ .
Thus, the oxidation number of Cl changes by $6$ .
$Equivalent weight = \frac{Molecular weight}{Change in the oxidation number}$
Hence, $E = \frac{M}{6}$

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