The equivalent weight of Na_2S_4O_6 in the re | vedclass

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(D) In the reaction $2Na_2S_2O_3 + I_2 	o Na_2S_4O_6 + 2NaI$,the oxidation state of sulfur in $Na_2S_2O_3$ is $+2$ and in $Na_2S_4O_6$ is $+2.5$.The

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$\frac{M}{2}$

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(D) In the reaction $2Na_2S_2O_3 + I_2 	o Na_2S_4O_6 + 2NaI$,the oxidation state of sulfur in $Na_2S_2O_3$ is $+2$ and in $Na_2S_4O_6$ is $+2.5$.The change in oxidation state per sulfur atom is $|2.5 - 2| = 0.5$.Since there are $4$ sulfur atoms in $Na_2S_4O_6$,the total $n$-factor is $0.5 	imes 4 = 2$.The equivalent weight $E$ is given by the formula $E = \frac{M}{n	ext{-factor}}$.Therefore,$E = \frac{M}{2}$.

The equivalent weight of $Na_2S_4O_6$ in the reaction $2Na_2S_2O_3 + I_2 \to Na_2S_4O_6 + 2NaI$ is :-

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