What will be the value of the equivalent weight of $K B r O_{3}$ in the following ionic equation?
$2 B r O_{3}^{-} + 12 H^{+} + 10 e^{-} \to B r_{2} + 6 H_{2} O$
(Given: The molecular weight of $K B r O_{3}$ is $M$ .)

\frac{M}{4}

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\frac{M}{6}

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\frac{M}{10}

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\frac{M}{5}

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Solution

The correct option is D $\frac{M}{5}$
$2 B r O_{3}^{-} + 12 H^{+} + 10 e^{-} \to B r_{2} + 6 H_{2} O$
$E_{K B r O_{3}} = \frac{molecular weight}{number of electrons gained} = \frac{2 B r O_{3}^{-}}{10} \Rightarrow \frac{B r O_{3}^{-}}{5} = \frac{K B r O_{3}}{5} = \frac{M}{5}$
Where $M$ stands for molecular weight of $K B r O_{3}$ .

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