Equivalent weight of `KMnO_(4)` when it is convert into `MnSO_(4)` is

Options

A. M/5
B. M/6
C. M/3
D. M/2

Explanation

To find the equivalent weight of \( KMnO_4 \) when it is converted into \( MnSO_4 \), we can follow these steps:

### Step 1: Determine the Molar Mass of \( KMnO_4 \)
The molar mass of \( KMnO_4 \) can be calculated by adding the atomic masses of its constituent elements:
- Potassium (K): 39.1 g/mol
- Manganese (Mn): 54.9 g/mol
- Oxygen (O): 16.0 g/mol (4 oxygen atoms)

Calculating the total:
\[ \text{Molar mass of } KMnO_4 = 39.1 + 54.9 + (4 \times 16.0) = 39.1 + 54.9 + 64.0 = 158.0 \text{ g/mol} \]

### Step 2: Determine the Change in Oxidation State of Manganese
In \( KMnO_4 \), manganese (Mn) is in the +7 oxidation state. In \( MnSO_4 \), manganese is in the +2 oxidation state. The change in oxidation state is:
\[ +7 \text{ to } +2 = 5 \text{ electrons} \]
This means that 5 electrons are gained when \( KMnO_4 \) is reduced to \( MnSO_4 \).

### Step 3: Calculate the Equivalent Weight
The equivalent weight can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{Number of electrons transferred}} \]
Substituting the values we have:
\[ \text{Equivalent weight of } KMnO_4 = \frac{158.0 \text{ g/mol}}{5} = 31.6 \text{ g/equiv} \]

### Conclusion
The equivalent weight of \( KMnO_4 \) when it is converted into \( MnSO_4 \) is \( 31.6 \text{ g/equiv} \).

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