Calculate the volume of 0.1N potassium permanganate that can be decolourised by 22.5 mL of 0.25 N ferrous ammonium sulphate solution in acidic medium.

Answer

To solve the problem of calculating the volume of 0.1 N potassium permanganate (KMnO4) that can be decolorized by 22.5 mL of 0.25 N ferrous ammonium sulfate (Fe(NH4)2(SO4)2) solution in acidic medium, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: In acidic medium, potassium permanganate (KMnO4) acts as an oxidizing agent and is reduced to Mn²⁺, while ferrous ions (Fe²⁺) are oxidized to ferric ions (Fe³⁺). The half-reactions can be summarized as:
- KMnO4 + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O
- Fe²⁺ → Fe³⁺ + e⁻

2. **Determine the Stoichiometry**: From the half-reactions, we see that 1 mole of KMnO4 reacts with 5 moles of Fe²⁺. Therefore, the stoichiometric ratio is:
- 1 mole of KMnO4 : 5 moles of Fe²⁺

3. **Calculate the Equivalent of Fe²⁺**: Since we have the normality (N) of the ferrous ammonium sulfate solution, we can calculate the equivalents of Fe²⁺ in the solution:
\[ \text{Equivalents of Fe²⁺} = \text{Normality} \times \text{Volume (L)} = 0.25 \, \text{N} \times 0.0225 \, \text{L} = 0.005625 \, \text{equivalents} \]

4. **Calculate the Equivalents of KMnO4 Required**: Using the stoichiometry, we know that 1 equivalent of KMnO4 reacts with 5 equivalents of Fe²⁺. Therefore, the equivalents of KMnO4 required are:
\[ \text{Equivalents of KMnO4} = \frac{\text{Equivalents of Fe²⁺}}{5} = \frac{0.005625}{5} = 0.001125 \, \text{equivalents} \]

5. **Calculate the Volume of KMnO4 Solution**: Now, we can calculate the volume of the KMnO4 solution needed using its normality:
\[ \text{Volume of KMnO4 (L)} = \frac{\text{Equivalents of KMnO4}}{\text{Normality of KMnO4}} = \frac{0.001125}{0.1} = 0.01125 \, \text{L} \]
Converting this to mL:
\[ \text{Volume of KMnO4 (mL)} = 0.01125 \times 1000 = 11.25 \, \text{mL} \]

### Final Answer:
The volume of 0.1 N potassium permanganate that can be decolorized by 22.5 mL of 0.25 N ferrous ammonium sulfate solution in acidic medium is **11.25 mL**.