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$C l_{2} + O H^{-} \to C l^{-} + C l O_{3}^{-}$ In the reaction, the equivalent weight of chlorine is [M = Mol.wt of chlorine]

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$\frac{3 M}{4}$

$\frac{3 M}{5}$

$\frac{M}{2}$

$M$

answer is B.

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Detailed Solution

$C l_{2} + \underset{}{\to} 2 \bar{C l}$
$C l_{2} + \overset{}{\to} 2 C l O_{3}^{-} + 10 \bar{e}$
$∴ (n - f) = \frac{2 \times 10}{2 + 10} = \frac{20}{12} = \frac{5}{3}$
$∴$ eq. wt = $\frac{3 M}{5}$

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