In the reaction : ${Na}_{2} S_{2} O_{3} + 4 {Cl}_{2} + 5 H_{2} O \overset{}{\to} {Na}_{2} {SO}_{4} + H_{2} {SO}_{4} + 8 HCl,$ the equivalent weight of ${Na}_{2} S_{2} O_{3}$ will be : (M = molecular weight of ${Na}_{2} S_{2} O_{3}$ )

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M/1

M/4

M/2

M/8

answer is B.

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Detailed Solution

$\begin{array}{l} \overset{+ 2}{{Na}_{2} S_{2} O_{3}} \overset{}{\to} \overset{+ 6}{{Na}_{2} {SO}_{4}} \\ the total change in oxidation number = 4 \times 2 = 8 \\ ∴ E_{{Na}_{2} S_{2} O_{3}} = \frac{mol . wt .}{v . f} = \frac{M}{8} \end{array}$

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