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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)^(-)` |
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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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