Calculate the volume of 0.05 M `KMnO_(4)` solution required to oxidise completely 2.70 grams of oxalic acid `(H_(2)C_(2)O_(4))` in acidic medium

Explanation

To solve the problem of calculating the volume of 0.05 M KMnO₄ solution required to completely oxidize 2.70 grams of oxalic acid (H₂C₂O₄) in an acidic medium, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Molar Mass of Oxalic Acid (H₂C₂O₄)**:
- The molar mass of H₂C₂O₄ can be calculated as follows:
- H: 1 g/mol × 2 = 2 g/mol
- C: 12 g/mol × 2 = 24 g/mol
- O: 16 g/mol × 4 = 64 g/mol
- Total molar mass = 2 + 24 + 64 = 90 g/mol

2. **Calculate the Number of Moles of Oxalic Acid**:
- Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
- For 2.70 g of oxalic acid:
\[ \text{Number of moles} = \frac{2.70 \text{ g}}{90 \text{ g/mol}} = 0.03 \text{ moles} \]

3. **Determine the Equivalent Factor (n-factor) for Oxalic Acid**:
- In acidic medium, oxalic acid (C₂O₄²⁻) is oxidized to CO₂.
- The change in oxidation state for carbon is from +3 to +4 (2 carbon atoms), so:
- n-factor for oxalic acid = 2 (number of moles of electrons transferred per mole of oxalic acid).

4. **Calculate the Equivalent of Oxalic Acid**:
- Equivalent = Number of moles × n-factor:
\[ \text{Equivalent of oxalic acid} = 0.03 \text{ moles} \times 2 = 0.06 \text{ equivalents} \]

5. **Determine the n-factor for KMnO₄**:
- In acidic medium, KMnO₄ (MnO₄⁻) is reduced to Mn²⁺.
- The change in oxidation state for manganese is from +7 to +2, which involves the transfer of 5 electrons.
- Therefore, n-factor for KMnO₄ = 5.

6. **Calculate the Volume of KMnO₄ Solution Required**:
- Using the formula:
\[ \text{Volume (L)} = \frac{\text{Equivalent of oxalic acid}}{\text{Molarity of KMnO₄} \times \text{n-factor of KMnO₄}} \]
- Rearranging gives:
\[ \text{Volume (L)} = \frac{0.06 \text{ equivalents}}{0.05 \text{ M} \times 5} \]
- Simplifying:
\[ \text{Volume (L)} = \frac{0.06}{0.25} = 0.24 \text{ L} \]
- Converting to mL:
\[ \text{Volume (mL)} = 0.24 \text{ L} \times 1000 = 240 \text{ mL} \]

### Final Answer:
The volume of 0.05 M KMnO₄ solution required to oxidize completely 2.70 grams of oxalic acid in acidic medium is **240 mL**.