If the molecular wt. of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{I}_2$ are $\mathrm{M}_1$ and $\mathrm{M}_2$ respectively, then what will be the equivalent wt. of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{I}_2$ in the following reaction? $2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \longrightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}$

Question

If the molecular wt. of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{I}_2$ are $\mathrm{M}_1$ and $\mathrm{M}_2$ respectively, then what will be the equivalent wt. of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ and $\mathrm{I}_2$ in the following reaction? $2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \longrightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}$, $\mathrm{M}_1, \mathrm{M}_2$, $\mathrm{M}_1, \mathrm{M}_2 / 2$, $2 \mathrm{M}_1, \mathrm{M}_2$, $\mathrm{M}_1, 2 \mathrm{M}_2$

MARKS App · Accepted Answer

Hints: $\begin{aligned} & \text { n.f. }\left(\mathrm{S}_2 \mathrm{O}_3^{2-}\right)=1, \text { Equivalent mass }=\frac{\mathrm{M}_1}{1}=\mathrm{M}_1 \ & \mathrm{n} \mathrm{f} .\left(\mathrm{I}_2\right)=2, \quad \text { Equivalent mass }=\frac{\mathrm{M}_2}{2} \end{aligned}$

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