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

Explanation

To find the equivalent weight of \( KMnO_4 \) when it oxidizes \( I^- \) to \( I_2 \) in acidic medium, we can follow these steps:

### Step 1: Identify the oxidation states
In the reaction, \( KMnO_4 \) acts as an oxidizing agent. We need to determine the change in oxidation states of manganese (Mn) in \( KMnO_4 \). In \( KMnO_4 \), manganese is in the +7 oxidation state.

### Step 2: Determine the final oxidation state
In acidic medium, \( MnO_4^- \) is reduced to \( Mn^{2+} \), where manganese is in the +2 oxidation state.

### Step 3: Calculate the change in oxidation state
The change in oxidation state for manganese is:
\[ \text{Change in oxidation state} = +7 - (+2) = 5 \]
This means that 1 mole of \( KMnO_4 \) can accept 5 moles of electrons.

### Step 4: Find the molecular weight of \( KMnO_4 \)
The molecular weight (gram molecular mass, GMM) of \( KMnO_4 \) is given as 158 g/mol.

### Step 5: 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 we found to be 5).

Substituting the values:
\[ \text{Equivalent weight} = \frac{158 \, \text{g/mol}}{5} = 31.6 \, \text{g/equiv} \]

### Final Answer
Thus, the equivalent weight of \( KMnO_4 \) is **31.6 g/equiv**.

---