1.

The equivalent mass of `MnSO_(4)` is half its molecular mass when it is converted toA. `MnO_(4)^(2-)`B. `Mn_(O_(3))`C. `MnO_(2)`D. `MnO_(1)^(-)`

Answer» Correct Answer - 3
`overset(+2)(M)nSO_(4) rarr overset(+6)(M)nO_(4)^(2-)`
`overset(+2)(M)nSO_(4) rarr overset(+3)(M)n_(2)O_(3)`
`overset(+2)(M)nSO_(4) rarr overset(+4)(M)nO_(2)`
`overset(+2)(M)nSO_(4) rarr overset(+6)(M)nO_(4)^(-)`
Only in `3^(rd)` option, change in oxidation number `(ON)` of `Mn` per formula unit of `MnSO_(4)=2`. Thus,
Equivalent mass of `MnSO_(4)=("Formula mass")/2`


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