A solution of ferrous oxalate has been prepared by dissolving 3.6 g `L^(-1)` calculate the volume of 0.01 M `KMnO_(4)` solution required for complete oxidation of 100 mL of ferrous oxalate solution in acidic medium

Explanation

To solve the problem, we need to determine the volume of 0.01 M KMnO₄ solution required for the complete oxidation of 100 mL of a ferrous oxalate solution that has a concentration of 3.6 g/L.

### Step-by-Step Solution:

1. **Calculate the Molarity of Ferrous Oxalate (C₂O₄²⁻):**
- The molar mass of ferrous oxalate (FeC₂O₄) can be calculated as follows:
- Molar mass of Fe = 55.85 g/mol
- Molar mass of C = 12.01 g/mol (2 x 12.01 = 24.02 g/mol for 2 carbon atoms)
- Molar mass of O = 16.00 g/mol (4 x 16.00 = 64.00 g/mol for 4 oxygen atoms)
- Total molar mass = 55.85 + 24.02 + 64.00 = 143.87 g/mol (approximately 144 g/mol)
- Given concentration = 3.6 g/L
- Molarity (M) = mass (g) / molar mass (g/mol) = 3.6 g / 144 g/mol = 0.025 M

2. **Determine the Equivalent Factor (n-factor) for Ferrous Oxalate:**
- The oxidation of ferrous ion (Fe²⁺) to ferric ion (Fe³⁺) involves a change of 1 electron.
- The n-factor for C₂O₄²⁻ when oxidized to CO₂ is 2 (since it loses 2 electrons).
- Therefore, the total n-factor for ferrous oxalate is 1 (for Fe²⁺) + 2 (for C₂O₄²⁻) = 3.

3. **Calculate the Number of Equivalents of Ferrous Oxalate in 100 mL:**
- Volume of solution = 100 mL = 0.1 L
- Number of equivalents (n) = Molarity (M) × Volume (L) × n-factor
- n = 0.025 M × 0.1 L × 3 = 0.0075 equivalents

4. **Determine the n-factor for KMnO₄:**
- In acidic medium, KMnO₄ (MnO₄⁻) is reduced from Mn⁷⁺ to Mn²⁺, which involves a change of 5 electrons.
- Therefore, the n-factor for KMnO₄ in acidic medium is 5.

5. **Use the Equivalent Relationship to Find the Volume of KMnO₄ Required:**
- Using the formula: n₁V₁ = n₂V₂, where:
- n₁ = number of equivalents of ferrous oxalate = 0.0075
- V₁ = volume of ferrous oxalate solution = 0.1 L
- n₂ = n-factor for KMnO₄ = 5
- V₂ = volume of KMnO₄ solution required
- Rearranging gives: V₂ = (n₁V₁) / n₂
- V₂ = (0.0075 equivalents × 0.1 L) / 5 = 0.00015 L = 0.15 mL

6. **Convert Volume to mL:**
- Since we need the volume in mL, we multiply by 1000:
- Volume of KMnO₄ required = 0.15 L × 1000 = 150 mL

### Final Answer:
The volume of 0.01 M KMnO₄ solution required for the complete oxidation of 100 mL of ferrous oxalate solution is **150 mL**.