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

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Correct Answer - B
Eq. mass = `("Molecular weight")/("Change in oxidation no of Mn")=("Mol.wt.")/(4-2)=("Mol. Wt.")/(2)`
(O.N. of Mn in `MnSO_(4)=+2`,O.N. of Mn in `MnO_(2)=+4`).

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