The equivalent weight of `MnSO_(4)` is half its molecular weight when it is converted to

Explanation

Equivalent weight = molecular weight / n-factor
If n factor is 1, then equivalent weight will be equal to its molecular weight
In `MnSO_(4)` the oxidation state of Mn is II , In `Mn_(2)O_(3)` the oxidation state of Mn is +III
In `MnO_(2)` the oxidation state of Mn is +IV,In `MnO_(4)^(-)` the oxidation state of Mn is +VII
In `MnO_(4)^(2-)` the oxidation state of Mn is +VI
Thus, when `MnSO_(4)` is converted into `MnO_(2)` then the n factor is 2, and the equivalent weight of `MnSO_(4)` will be half of its molecular weight