Question

In the reaction $I_2+2S_2{O}_3^{2-}\to S_4{O}_6^{2-}+2I^{-}$, the equivalent weight of iodine will be equal to:

• A. its molecular weight
• B. $\frac{1}{2}$ of its molecular weight
• C. $\frac{1}{4}$ of its molecular weight
• D. twice of its molecular weight

Answer 1

The correct option is B $\frac{1}{2}$ of its molecular weight

$Equivalentweight=\frac{Molecularweight}{Valencyfactor}$

$2S_2{O}_3^{-2}+I_2^0\to 2I^{-}+S_4{O}_6^{-2}$

$Valencyfactor=2\times 1=2$

$\therefore Equivalentweight=\frac{Molecularweight}{2}$