The equivalent weight of MnSO_4 is half its m | vedclass

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

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$MnO_2$

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(B) The equivalent weight of a substance is given by the formula: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.For the equivalent weight to be half the molecular weight,the n-factor must be $2$.The n-factor represents the change in oxidation state of the central metal atom.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,for the conversion $MnSO_4 	o MnO_2$,the n-factor is $2$,making the equivalent weight $\frac{M}{2}$.

List $I$ (Compound)	List $II$ (Oxidation state of $N$)
$(A)$ $NO_2$	$(1)$ $+5$
$(B)$ $HNO$	$(2)$ $-3$
$(C)$ $NH_3$	$(3)$ $+4$
$(D)$ $N_2O_5$	$(4)$ $+1$

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

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