`KMnO_(4)` oxidises `I^(-)` to `I_(2)` in acidic medium. The equivalent weight of `KMnO_(4)` is :

Explanation

To find the equivalent weight of `KMnO4` when it oxidizes `I^(-)` to `I2` in acidic medium, we can follow these steps:

### Step 1: Write the balanced reaction
The skeletal reaction can be written as:
\[ \text{MnO}_4^{-} + \text{I}^{-} \rightarrow \text{Mn}^{2+} + \text{I}_2 \]

### Step 2: Determine the change in oxidation state of Manganese
In `KMnO4`, the oxidation state of manganese (Mn) is +7. In the product `Mn^2+`, the oxidation state of manganese is +2.
- Change in oxidation state of Mn:
\[ +7 \text{ (in KMnO}_4) \rightarrow +2 \text{ (in Mn}^{2+}) \]
- The change in oxidation state is:
\[ 7 - 2 = 5 \]

### Step 3: Calculate the equivalent weight
The equivalent weight is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} \]
where \( n \) is the number of electrons transferred (which is equal to the change in oxidation state).

### Step 4: Calculate 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)

Thus, the molecular weight of `KMnO4` is:
\[ 39 + 55 + (16 \times 4) = 39 + 55 + 64 = 158 \text{ g/mol} \]

### Step 5: Substitute into the equivalent weight formula
Now substituting the values into the equivalent weight formula:
\[ \text{Equivalent weight} = \frac{158 \text{ g/mol}}{5} = 31.6 \text{ g/equiv} \]

### Conclusion
The equivalent weight of `KMnO4` is **31.6 g/equiv**.

---