In strongly alkaline medium, `MnO_4^-`
`MnO_4^(-) + e^(-) to MnO_4^(2-)`
In this medium, equivalent weight of `KMnO_4` is reduced as:

Options

A. 31.6
B. 52.67
C. 79
D. 158

Explanation

To determine the equivalent weight of KMnO4 in a strongly alkaline medium where the reduction of MnO4^- to MnO4^(2-) occurs, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction in strongly alkaline medium is:
\[ \text{MnO}_4^- + e^- \rightarrow \text{MnO}_4^{2-} \]

2. **Determine the Oxidation States**:
In MnO4^-, manganese (Mn) has an oxidation state of +7. In MnO4^(2-), the oxidation state of manganese is +6. This indicates that manganese is being reduced.

3. **Calculate the Change in Oxidation State**:
The change in oxidation state from +7 to +6 corresponds to the gain of 1 electron. Therefore, the n-factor (the number of electrons gained or lost per formula unit) for this reaction is 1.

4. **Find the Molecular Weight of KMnO4**:
The molecular weight of KMnO4 can be calculated as follows:
- Potassium (K) = 39 g/mol
- Manganese (Mn) = 55 g/mol
- Oxygen (O) = 16 g/mol (and there are 4 oxygen atoms)
\[ \text{Molecular Weight of KMnO}_4 = 39 + 55 + (4 \times 16) = 39 + 55 + 64 = 158 \text{ g/mol} \]

5. **Calculate the Equivalent Weight**:
The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} \]
Substituting the values we have:
\[ \text{Equivalent Weight} = \frac{158 \text{ g/mol}}{1} = 158 \text{ g/equiv} \]

6. **Conclusion**:
Therefore, the equivalent weight of KMnO4 in a strongly alkaline medium is 158 g/equiv.

### Final Answer:
The equivalent weight of KMnO4 is **158 g/equiv**.