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 :

Options

A. 40 mL
B. 20 mL
C. 10 mL
D. 4 mL

Explanation

To solve the problem step by step, we will follow the outlined procedure to determine the volume of 0.1 N NaOH required to completely neutralize 10 mL of the oxalic acid dihydrate solution.

### Step 1: Calculate the molar mass of oxalic acid dihydrate (C2H2O4·2H2O).
- The molecular formula of oxalic acid dihydrate is C2H2O4·2H2O.
- Calculate the molar mass:
- Carbon (C): 2 × 12 g/mol = 24 g/mol
- Hydrogen (H): 2 × 1 g/mol = 2 g/mol
- Oxygen (O): 4 × 16 g/mol = 64 g/mol
- Water (H2O): 2 × (2 × 1 + 16) = 36 g/mol
- Total molar mass = 24 + 2 + 64 + 36 = 126 g/mol

### Step 2: Determine the basicity of oxalic acid.
- Oxalic acid has 2 acidic protons (H), so its basicity = 2.

### Step 3: Calculate the equivalent weight of oxalic acid dihydrate.
- Equivalent weight = Molar mass / Basicity = 126 g/mol / 2 = 63 g/equiv.

### Step 4: Calculate the number of equivalents in the given mass of oxalic acid.
- Given mass = 6.3 g
- Number of equivalents = Given mass / Equivalent weight = 6.3 g / 63 g/equiv = 0.1 equiv.

### Step 5: Calculate the normality of the oxalic acid solution.
- The solution is made up to 250 mL (0.250 L).
- Normality (N) = Number of equivalents / Volume of solution in liters = 0.1 equiv / 0.250 L = 0.4 N.

### Step 6: Use the normality equation to find the volume of NaOH required for neutralization.
- We need to neutralize 10 mL of the oxalic acid solution.
- Using the equation: N1V1 = N2V2
- Where:
- N1 = Normality of oxalic acid = 0.4 N
- V1 = Volume of oxalic acid solution = 10 mL
- N2 = Normality of NaOH = 0.1 N
- V2 = Volume of NaOH required (unknown)
- Substitute the values into the equation:
- 0.4 N × 10 mL = 0.1 N × V2
- Rearranging gives:
- V2 = (0.4 N × 10 mL) / 0.1 N = 40 mL

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

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