The equivalent weight of chlorine molecule in the given equation is
$3 C l_{2} + 6 N a O H \to 5 N a C l + N a C l O_{3} + 3 H_{2} O$

42.6

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35.5

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59.1

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71

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Solution

The correct option is A $42.6$
Oxidation half: $6 O H^{-} + \frac{1}{2} C l_{2} \to C l O_{3}^{-} + 5 e^{-} + 3 H_{2} O$
Reduction half: $\frac{5}{2} C l_{2} + 5 e^{-} \to 5 C l^{-}$
As we can see from above, there is a change of $5$ electrons in the redox reaction.
Now, if we add the oxidation half and reduction half, we will get $3$ moles of $C l_{2}$ .
$\frac{1}{2} C l_{2}$ (Oxidation) + $\frac{5}{2} C l_{2}$ (Reduction) $= \frac{6}{2} = 3$ moles of $C l_{2}$
Now molar mass of elemental chlorine = 35.5 amu
Therefore, molar mass of chlorine gas, $C l_{2} = 2 \times 35.5 a m u = 71 a m u$
$equivalent mass of an oxidizing or reducing agent = \frac{moles of oxidising or reducing agent. x molar mass}{number of electrons taking part in the reaction}$
Therefore, equivalent mass of $C l_{2} = \frac{3 \times 71}{5} = 42.6$

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