The equivalent weight of MnSO_4 is half of it | vedclass

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(B) The equivalent weight is given by the formula: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{n	ext{-factor}}$.For the equivalent wei

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$MnO_2$

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(B) The equivalent weight is given by the formula: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{n	ext{-factor}}$.For the equivalent weight to be half of the molecular weight,the $n$-factor must be $2$.In $MnSO_4$,the oxidation state of $Mn$ is $+2$.In $MnO_2$,the oxidation state of $Mn$ is $+4$.The change in oxidation state is $|4 - 2| = 2$.Thus,the reaction $Mn^{2+} ightarrow Mn^{4+} + 2e^-$ involves the loss of $2$ electrons,making the $n$-factor $2$.

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

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