Question

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

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

Answer 1

The correct option is B $Mn{O}_2$

$\stackrel{+2}{Mn}S{O}_4^2\to \stackrel{+4}{Mn}{O}_2+2e^{-}$

Here, change in the oxidation of Mn = 2.

So, n−factor =2.

Equivalent weight $=\frac{M}{2}$

Therefore, the equivalent weight of MnSO4 is half of its molecular weight when it is converted to $Mn{O}_2$.