If the molecular weights of Na_2S_2O_3 and I_ | vedclass

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(B) The given reaction is: $2S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2I^-$For $S_2O_3^{2-}$: The oxidation state of $S$ changes from $+2$ to $

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$M_1, M_2 / 2$

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(B) The given reaction is: $2S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2I^-$For $S_2O_3^{2-}$: The oxidation state of $S$ changes from $+2$ to $+2.5$. The change in oxidation number per $S$ atom is $0.5$. Since there are $2$ sulfur atoms in $S_2O_3^{2-}$, the total change per mole is $0.5 	imes 2 = 1$. Thus, the $n$-factor is $1$. Equivalent weight = $\frac{M_1}{1} = M_1$.For $I_2$: The oxidation state of $I$ changes from $0$ to $-1$. The change in oxidation number per $I$ atom is $1$. Since there are $2$ iodine atoms in $I_2$, the total change per mole is $1 	imes 2 = 2$. Thus, the $n$-factor is $2$. Equivalent weight = $\frac{M_2}{2}$.

If the molecular weights of $Na_2S_2O_3$ and $I_2$ are $M_1$ and $M_2$ respectively, then what will be the equivalent weights of $Na_2S_2O_3$ and $I_2$ in the following reaction?
$2S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2I^-$

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If the molecular weights of $Na_2S_2O_3$ and $I_2$ are $M_1$ and $M_2$ respectively, then what will be the equivalent weights of $Na_2S_2O_3$ and $I_2$ in the following reaction?$2S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2I^-$