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  4. The equivalent weight of MnSO4 is half of its molecular weight, when it converts to

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Q. The equivalent weight of $MnSO_4$ is half of its molecular
weight, when it converts to

IIT JEEIIT JEE 1988Some Basic Concepts of Chemistry

Solution:

Equivalent weight in redox system is defined as :
$E =\frac{\text{Molar mass}}{\text{n-factor}}$
Here $n$-factor is the net change in oxidation number per formula unit of oxidising or reducing agent. In the present case, $n$-factoris $2$ because equivalent weight is half of molecular weight. Also,
$n$-factor $MnSO_4 \rightarrow \, \frac{1}{2}Mn_2O_3 \, \, 1(+2 \rightarrow \, +3)$
$ MnSO_4 \rightarrow MnO_2 \, \, 2(+2 \rightarrow +4)$
$ MnSO_4 \rightarrow MnO_4^- \, \, 5(+2 \rightarrow +7)$
$ MnSO_4 \rightarrow MnO_4^2- \, \, 4(+2 \rightarrow +6)$
Therefore , MnS$O_4$ converts to $MnO_2$