1.17 g an impure sample of oxalic acid dihydrate was dissolved and make up of 200 ml with water. 10 ml of this solution in acidic medium required 8.5 ml of a solution of potassium permanganate containing 3.16 g per litre of solution. The percentage purity of oxalic acid will be:

Options

A. 0.1265
B. 0.3578
C. 0.8276
D. 0.9154

Answer

To find the percentage purity of oxalic acid dihydrate in the given sample, we can follow these steps:

### Step 1: Calculate the equivalent weight of oxalic acid dihydrate (H2C2O4·2H2O)
1. **Molar Mass Calculation**:
- Molar mass of H2C2O4·2H2O:
- H: 2 × 1 = 2 g/mol
- C: 2 × 12 = 24 g/mol
- O: 4 × 16 = 64 g/mol
- H2O: 2 × (2 × 1 + 16) = 36 g/mol
- Total Molar Mass = 2 + 24 + 64 + 36 = 126 g/mol

2. **Equivalent Weight Calculation**:
- n-factor for oxalic acid (as it can donate 2 H+ ions): 2
- Equivalent weight = Molar mass / n-factor = 126 g/mol / 2 = 63 g/equiv

### Step 2: Calculate the normality of the potassium permanganate (KMnO4) solution
1. **Given**: 3.16 g of KMnO4 per liter
2. **Molar Mass of KMnO4**:
- K: 39 g/mol
- Mn: 54.9 g/mol
- O: 4 × 16 = 64 g/mol
- Total Molar Mass = 39 + 54.9 + 64 = 157.9 g/mol (approximately 158 g/mol)

3. **n-factor for KMnO4**:
- In acidic medium, KMnO4 gains 5 electrons (n-factor = 5).

4. **Equivalent Weight of KMnO4**:
- Equivalent weight = Molar mass / n-factor = 158 g/mol / 5 = 31.6 g/equiv

5. **Normality Calculation**:
- Normality (N) = Weight (g) / (Equivalent weight × Volume in L)
- For 3.16 g in 1 L: N = 3.16 g / (31.6 g/equiv) = 0.1 N

### Step 3: Calculate the equivalents of KMnO4 used in the reaction
1. **Volume of KMnO4 solution used**: 8.5 mL = 0.0085 L
2. **Equivalents of KMnO4**:
- Equivalents = Normality × Volume in L = 0.1 N × 0.0085 L = 0.00085 equivalents

### Step 4: Relate the equivalents of KMnO4 to the equivalents of oxalic acid
1. **From the reaction**: 1 equivalent of KMnO4 reacts with 1 equivalent of oxalic acid.
2. **Equivalents of oxalic acid** = 0.00085 equivalents.

### Step 5: Calculate the weight of pure oxalic acid in the sample
1. **Weight of oxalic acid** = Equivalents × Equivalent weight
- Weight = 0.00085 equivalents × 63 g/equiv = 0.05355 g (or 0.0536 g when rounded).

### Step 6: Calculate the percentage purity of the oxalic acid sample
1. **Percentage purity** = (Weight of pure oxalic acid / Weight of impure sample) × 100
- Percentage purity = (0.0536 g / 1.17 g) × 100 = 4.58%

### Final Answer
The percentage purity of the oxalic acid dihydrate sample is approximately **4.58%**.