In acidic medium, the equivalent weight of `KMnO_(4)` will be

Options

A. 31.6
B. 52.7
C. `7.0`
D. `158.0`

Explanation

To find the equivalent weight of \( KMnO_4 \) in acidic medium, 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 g/mol
- Manganese (Mn): 55 g/mol
- Oxygen (O): 16 g/mol (4 oxygen atoms)

Calculating the molar mass:
\[ \text{Molar mass of } KMnO_4 = 39 + 55 + (4 \times 16) = 39 + 55 + 64 = 158 \text{ g/mol} \]

### Step 2: Determine the Change in Oxidation State of Manganese
In \( KMnO_4 \), manganese (Mn) is in the +7 oxidation state. In acidic medium, it gets reduced to Mn\(^{2+}\), which has an oxidation state of +2.

Calculating the change in oxidation state:
\[ \text{Change in oxidation state} = +7 - (+2) = 5 \]

### Step 3: Determine the n Factor
The n factor is defined as the total change in oxidation number per formula unit of the substance. Since there is one manganese atom undergoing this change:
\[ \text{n factor} = \text{Change in oxidation state} \times \text{Number of atoms undergoing change} = 5 \times 1 = 5 \]

### Step 4: Calculate the Equivalent Weight
The equivalent weight is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n factor}} \]
Substituting the values we have:
\[ \text{Equivalent weight} = \frac{158 \text{ g/mol}}{5} = 31.6 \text{ g/equiv} \]

### Conclusion
Thus, the equivalent weight of \( KMnO_4 \) in acidic medium is **31.6 g/equiv**.

---