The equivalent weight of `MnSO_4` is half its molecular weight when it is com1erted to (a) `Mn_(2) O_(3)` (b) `MnO_(2)` (c ) `MnO_(4)^(-)` (d) `MnO_(4)^(2-)`. Indicate the correct answer.

Explanation

To determine which conversion of `MnSO4` results in an equivalent weight that is half of its molecular weight, we need to analyze the n-factor for each conversion. The n-factor is the number of moles of electrons transferred per mole of the substance in a redox reaction.

1. **Calculate the Molecular Weight of `MnSO4`:**
- Molar mass of Mn = 54.94 g/mol
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol
- Therefore, the molecular weight of `MnSO4` = 54.94 + 32.07 + (4 × 16.00) = 54.94 + 32.07 + 64.00 = 150.01 g/mol.

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