The equivalent weight of Na_2S_2O_3 (Gram mol | vedclass

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(B) In the given reaction: $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$. The oxidation state of sulfur in $Na_2S_2O_3$ changes from $+2$ to $+2.5

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$M$

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(B) In the given reaction: $I_2 + 2Na_2S_2O_3 ightarrow 2NaI + Na_2S_4O_6$. The oxidation state of sulfur in $Na_2S_2O_3$ changes from $+2$ to $+2.5$ in $Na_2S_4O_6$. The change in oxidation state per sulfur atom is $2.5 - 2 = 0.5$. Since there are $2$ sulfur atoms in $Na_2S_2O_3$, the total change in oxidation state per molecule is $2 	imes 0.5 = 1$. Thus, the $n$-factor for $Na_2S_2O_3$ is $1$. Equivalent weight $= \frac{	ext{Molecular weight}}{n	ext{-factor}} = \frac{M}{1} = M$.

The equivalent weight of $Na_2S_2O_3$ (Gram molecular weight $= M$) in the given reaction is $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.

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