Question

The equivalent weight of $MnS{O}_4$ is half of its molecular weight when it converts to :

• A. $M{n}_2{O}_3$
• B. $Mn{O}_2$
• C. $Mn{O}_4^{-}$
• D. $Mn{O}_4^{2-}$

Answer 1

Answer: (B)

$Mn{O}_2$

The expression for the equivalent weight is $E=\frac{Molarmass}{n-factor}$.
Here, n- factor represents the net change in oxidation state per formula unit oxidant or reductant.
When $MnS{O}_4$ converts to $Mn{O}_2$, the oxidation state changes from +2 to +4.
Hence, the value of n-factor is 2.
The equivalent weight $E=\frac{Molarmass}{n-factor}=\frac{Molarmass}{2}$.