The equivalent weight of hypo in the reaction [M = molecular weight]
2Na₂S₂O₃ + I₂ → 2NaI + Na₂S₄O₆ is

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M/4

M/3

M/2

answer is A.

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Detailed Solution

$2 {Na}_{2} {\overset{- 2}{S}}_{2} O_{3} + I_{2} \to 2 NaI + {Na}_{2} {\overset{2.5}{S}}_{4} O_{6} n - factor of {Na}_{2} S_{2} O_{3} = 2 (2.5 - 2) = 1$

Equivalent weight is M/1.

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