A : Equivalent weight of `KMnO_(4)` in acidic medium is `M/5`.
R : In acidic medium 1 mol of `MnO_(4)""^(-)` gains 5 electron.

Options

A. If both Assertion & Reason are true and the reason is the correct explanation of the assertion, then mark (1).
B. If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2).
C. If Assertion is true statement but Reason is false, then mark (3).
D. If both Assertion and Reason are false statements, then mark (4).

Answer

To solve the problem, we need to determine the equivalent weight of potassium permanganate (KMnO₄) in acidic medium and understand the reasoning behind it. Let's break it down step by step.

### Step 1: Understanding Equivalent Weight
The equivalent weight of a substance is defined as the mass of that substance that will combine with or displace 1 mole of hydrogen ions (H⁺) in a reaction. It can be calculated using the formula:

\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{n}} \]

where \( n \) is the number of electrons gained or lost in the reaction.

### Step 2: Molar Mass of KMnO₄
First, we need to calculate the molar mass of KMnO₄:
- Potassium (K) = 39.1 g/mol
- Manganese (Mn) = 54.9 g/mol
- Oxygen (O) = 16.0 g/mol × 4 = 64.0 g/mol

Adding these together gives:

\[ \text{Molar Mass of KMnO₄} = 39.1 + 54.9 + 64.0 = 158.0 \text{ g/mol} \]

### Step 3: Determining the Valency Factor
In acidic medium, potassium permanganate (KMnO₄) acts as an oxidizing agent. The permanganate ion (MnO₄⁻) is reduced to manganese ions (Mn²⁺).

The half-reaction for this reduction can be represented as:

\[ \text{MnO}_4^{-} + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]

From this half-reaction, we can see that 1 mole of MnO₄⁻ gains 5 electrons. Therefore, the valency factor \( n \) for KMnO₄ in acidic medium is 5.

### Step 4: Calculating Equivalent Weight
Now that we have the molar mass and the valency factor, we can calculate the equivalent weight:

\[ \text{Equivalent Weight of KMnO₄} = \frac{158.0 \text{ g/mol}}{5} = 31.6 \text{ g/equiv} \]

### Conclusion
Thus, the equivalent weight of KMnO₄ in acidic medium is \( \frac{M}{5} \), where \( M \) is the molar mass of KMnO₄.