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  4. The equivalent weight of MnSO4 is half its molecular weight when it is converted to:

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Q. The equivalent weight of $MnSO_4$ is half its molecular weight when it is converted to:

Some Basic Concepts of Chemistry

Solution:

Equivalent weight $=\frac{molecular \,weight}{valency\, factor}$
If valency factor is $2$, then equivalent weight will be equal to its molecular weight.
In $MnSO_4,$ the oxidation state of $Mn$ is $+II$
In $Mn_2O_3,$ 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 $MnO^{2-}_4$; the oxidation state of $Mn$ is $+VI$
Thus, when $MnSO_4$ is converted into $MnO_2,$ then the valency factor is $2$, and the equivalent weight of $MnSO_4$ will be half of its molecular weight.