An aq. Solution of 0.5 M `KMnO_(4)` is divided into two parts. One part of it requires 125mL of 1.5M aq. Solution of oxalate ions in acidic medium, while another part requires 270mL of 0.5 M aq. Solution of iodide ions in neutral medium which are converted into `I_(2)` only. Calculated total volume (mL) of intial `KMnO_(4)` solution.

Answer

To solve the problem, we need to determine the total volume of the initial KMnO4 solution based on its reactions with oxalate ions and iodide ions.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The first part of KMnO4 reacts with oxalate ions (C2O4^2-) in acidic medium.
- The second part reacts with iodide ions (I-) in neutral medium to produce I2.

2. **Calculate the Volume of KMnO4 for Oxalate Reaction**:
- The reaction for oxalate ions can be represented as:
\[ 2 \text{C}_2\text{O}_4^{2-} + 2 \text{MnO}_4^{-} + 16 \text{H}^+ \rightarrow 4 \text{CO}_2 + 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} \]
- Here, the N-factor for KMnO4 in acidic medium is 5 (Mn changes from +7 to +2).
- The N-factor for oxalate ions is 2 (each oxalate ion loses 2 electrons).

3. **Calculate Milliequivalents for Oxalate**:
- Volume of oxalate solution = 125 mL
- Concentration of oxalate solution = 1.5 M
- Milliequivalents of oxalate = Volume (mL) × Concentration (M) × N-factor
\[ \text{Milliequivalents of oxalate} = 125 \, \text{mL} \times 1.5 \, \text{M} \times 2 = 375 \, \text{meq} \]

4. **Set Up Equation for KMnO4**:
- Let \( V_1 \) be the volume of KMnO4 used for oxalate.
- The milliequivalents of KMnO4 = \( V_1 \times 0.5 \, \text{M} \times 5 \)
- Setting the milliequivalents equal:
\[ V_1 \times 0.5 \times 5 = 375 \]
\[ V_1 = \frac{375}{2.5} = 150 \, \text{mL} \]

5. **Calculate the Volume of KMnO4 for Iodide Reaction**:
- The reaction for iodide ions can be represented as:
\[ 2 \text{MnO}_4^{-} + 6 \text{I}^- + 8 \text{H}^+ \rightarrow 2 \text{MnO}_2 + 3 \text{I}_2 + 4 \text{H}_2\text{O} \]
- Here, the N-factor for KMnO4 in neutral medium is 3 (Mn changes from +7 to +4).
- The N-factor for iodide ions is 1 (each iodide ion loses 1 electron).

6. **Calculate Milliequivalents for Iodide**:
- Volume of iodide solution = 270 mL
- Concentration of iodide solution = 0.5 M
- Milliequivalents of iodide = Volume (mL) × Concentration (M) × N-factor
\[ \text{Milliequivalents of iodide} = 270 \, \text{mL} \times 0.5 \, \text{M} \times 1 = 135 \, \text{meq} \]

7. **Set Up Equation for KMnO4**:
- Let \( V_2 \) be the volume of KMnO4 used for iodide.
- The milliequivalents of KMnO4 = \( V_2 \times 0.5 \, \text{M} \times 3 \)
- Setting the milliequivalents equal:
\[ V_2 \times 0.5 \times 3 = 135 \]
\[ V_2 = \frac{135}{1.5} = 90 \, \text{mL} \]

8. **Calculate Total Volume of KMnO4**:
- Total volume of KMnO4 = \( V_1 + V_2 \)
\[ \text{Total Volume} = 150 \, \text{mL} + 90 \, \text{mL} = 240 \, \text{mL} \]

### Final Answer:
The total volume of the initial KMnO4 solution is **240 mL**.