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Q. In the reaction $2Na_2S_2O_3 + I_2 \to Na_2S_4O_6 + 2NaI$, the equivalent weight of $Na_2S_2O_3$(mol. wt. = $M$) is equal to

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Solution:

$Na_{2}^{+2}SO_3 +I_2 \to Na_{2}^{+2.5}S_4O_6+NaI$
$n = 2 \times 0.5 = 1$
$E=\frac{M}{n-factor}=\frac{M}{1}=M$