Equivalent weight of chlorine molecule in the equation is:
$3 C l_{2} + 6 N a O H ⟶ 5 N a C l + N a C l O_{3} + 3 H_{2} O$

42.6

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35.5

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59.1

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Solution

The correct option is A 42.6
$3 C l_{2} + 6 N a O H ⟶ 5 N a C l + N a C l O_{3} + 3 H_{2} O$
Oxidation: $6 O H^{-} + \frac{1}{2} C l_{2} ⟶ {2 C l O_{3}}^{-} + 5 e^{-} + 3 H_{2} O$
Reduction: $5 / 2 C l_{2} + 5 e^{-} ⟶ 5 C l^{-}$
$3$ moles of $C l_{2}$ & $5$ moles of electrons are involved in the redox reaction.
Molar mass of $C l_{2} = 71$
$∴$ Equivalent mass of $C l_{2} = \frac{3 \times 71}{5} = 42.6 g / m o l$

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