The equivalent weight of in the reaction is [M = molecular weight]: — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

In $Na_2S_2O_3$, the O.S. of $S$ is $+2$. In $Na_2S_4O_6$, the O.S. of $S$ is $+2.5$. <br>Change in O.S. per $S$ atom = $0.5$. <br>Since there are 2 $S$ atoms per molecule of $Na_2S_2O_3$, total electrons lost ($n$-factor) = $2 \times 0.5 = 1$. <br>Equivalent weight = $M / n = M / 1 = M$. [1-34]