In the reaction 2Na_2S_2O_3 + I_2 to Na_2S_4O | vedclass

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(A) 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 -

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$M$

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(A) 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 one molecule of $Na_2S_2O_3$,the total change in oxidation state (n-factor) is $2 	imes 0.5 = 1$.The equivalent weight $E$ is calculated as $E = \frac{	ext{Molecular Weight}}{	ext{n-factor}} = \frac{M}{1} = M$.

In the reaction $2Na_2S_2O_3 + I_2 \to Na_2S_4O_6 + 2NaI$,the equivalent weight of $Na_2S_2O_3$ (molecular weight = $M$) is equal to:

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