An aqueous solution of 6.3 g of oxalic acid dihydrate is made upto 250 mL. The volume of 0.1 N NaOH required to completely neutralise 10 mL of this solution is :

Explanation

To solve the problem, we need to determine the volume of 0.1 N NaOH required to completely neutralize 10 mL of a solution of oxalic acid dihydrate (H2C2O4·2H2O) prepared from 6.3 g of the acid in 250 mL of solution.

### Step-by-Step Solution:

1. **Determine the molar mass of oxalic acid dihydrate (H2C2O4·2H2O)**:
- Molar mass of H2C2O4 = (2 × 1) + (2 × 12) + (4 × 16) = 2 + 24 + 64 = 90 g/mol
- Molar mass of 2H2O = 2 × (2 × 1 + 16) = 2 × 18 = 36 g/mol
- Total molar mass of H2C2O4·2H2O = 90 + 36 = 126 g/mol

2. **Calculate the number of moles of oxalic acid in 6.3 g**:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{6.3 \text{ g}}{126 \text{ g/mol}} = 0.05 \text{ moles} \]

3. **Calculate the normality of the oxalic acid solution**:
- Oxalic acid (H2C2O4) can donate 2 protons (H+ ions), so its n-factor is 2.
\[ \text{Normality (N)} = \text{n-factor} \times \text{Molarity} \]
- First, we need to find the molarity:
\[ \text{Molarity} = \frac{\text{moles}}{\text{volume in L}} = \frac{0.05 \text{ moles}}{0.250 \text{ L}} = 0.2 \text{ M} \]
- Now, calculate the normality:
\[ \text{Normality} = 2 \times 0.2 = 0.4 \text{ N} \]

4. **Calculate the volume of 0.1 N NaOH required to neutralize 10 mL of the oxalic acid solution**:
- Use the formula for neutralization:
\[ N_1 V_1 = N_2 V_2 \]
- Where:
- \(N_1 = 0.4 \text{ N}\) (normality of oxalic acid solution)
- \(V_1 = 10 \text{ mL}\) (volume of oxalic acid solution)
- \(N_2 = 0.1 \text{ N}\) (normality of NaOH)
- \(V_2\) = volume of NaOH required (unknown)
- Plugging in the values:
\[ 0.4 \times 10 = 0.1 \times V_2 \]
\[ 4 = 0.1 \times V_2 \]
\[ V_2 = \frac{4}{0.1} = 40 \text{ mL} \]

### Final Answer:
The volume of 0.1 N NaOH required to completely neutralize 10 mL of the oxalic acid solution is **40 mL**.