In order to prepare one litre normal solution of `KMnO_(4)`, how many grams of `KMnO_(4)` are required if the solution is used in acidic medium for oxidation

Options

A. 158 g
B. 31.6 g
C. 79 g
D. 52.6 g

Answer

To prepare a 1-litre normal solution of KMnO₄ in acidic medium, we need to calculate the amount of KMnO₄ required in grams. Here’s a step-by-step solution:

### Step 1: Determine the Equivalent Weight of KMnO₄
In acidic medium, KMnO₄ acts as an oxidizing agent and the manganese (Mn) is reduced from +7 oxidation state to +2. The number of electrons transferred (n-factor) in this reaction is 5.

**Hint:** Remember that the n-factor is determined by the number of electrons gained or lost in a redox reaction.

### Step 2: 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 × 4 = 64 g/mol

So, the total molecular weight of KMnO₄ = 39 + 55 + 64 = 158 g/mol.

**Hint:** Always check the periodic table for the atomic weights of the elements when calculating molecular weight.

### Step 3: Calculate the Equivalent Weight
The equivalent weight can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n \text{-factor}} \]
Substituting the values:
\[ \text{Equivalent Weight} = \frac{158 \text{ g/mol}}{5} = 31.6 \text{ g/equiv} \]

**Hint:** The equivalent weight tells you how much of the substance is needed to provide one mole of reactive units (in this case, electrons).

### Step 4: Calculate the Amount Required for 1 Litre Normal Solution
Since we need a 1-litre normal solution, we require 1 equivalent of KMnO₄. Therefore, the mass required is equal to the equivalent weight:
\[ \text{Mass of KMnO₄ required} = \text{Equivalent Weight} = 31.6 \text{ grams} \]

**Hint:** For normal solutions, the number of equivalents is equal to the volume of the solution in litres.

### Conclusion
To prepare a 1-litre normal solution of KMnO₄ in acidic medium, you will need 31.6 grams of KMnO₄.

**Final Answer:** 31.6 grams of KMnO₄.