The equivalent weight of MnSO_4 is half of it | vedclass

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Question

(B) The equivalent weight $(E)$ is given by $E = \frac{M}{n}$,where $M$ is the molecular weight and $n$ is the change in oxidation state per molecule.

Vedclass · Accepted Answer

$MnO_2$

Vedclass · Answer

(B) The equivalent weight $(E)$ is given by $E = \frac{M}{n}$,where $M$ is the molecular weight and $n$ is the change in oxidation state per molecule.Given $E = \frac{M}{2}$,it implies $n = 2$.In $MnSO_4$,the oxidation state of $Mn$ is $+2$.For $MnO_2$,the oxidation state of $Mn$ is $+4$.Change in oxidation state = $|4 - 2| = 2$.Since the change in oxidation state is $2$,the equivalent weight is half of the molecular weight.

The equivalent weight of $MnSO_4$ is half of its molecular weight when it is converted into .....

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