What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N - NaOH solution

Options

A. 0.45 g
B. 0.90 g
C. 1.08 g
D. 1.26 g

Explanation

To find the weight of hydrated oxalic acid required for the complete neutralization of 100 ml of 0.2N NaOH solution, we can follow these steps:

### Step 1: Calculate the moles of NaOH
Given:
- Normality (N) of NaOH = 0.2 N
- Volume of NaOH solution = 100 ml = 0.1 L (since 1000 ml = 1 L)

Using the formula:
\[ \text{Moles of NaOH} = \text{Normality} \times \text{Volume in L} \]
\[ \text{Moles of NaOH} = 0.2 \, \text{N} \times 0.1 \, \text{L} = 0.02 \, \text{moles} \]

### Step 2: Determine the moles of oxalic acid needed
Oxalic acid (H₂C₂O₄) is a diprotic acid, meaning it can donate 2 protons (H⁺ ions). Therefore, the relationship between moles of NaOH and moles of oxalic acid is:
\[ \text{Moles of oxalic acid} = \frac{\text{Moles of NaOH}}{2} \]
\[ \text{Moles of oxalic acid} = \frac{0.02}{2} = 0.01 \, \text{moles} \]

### Step 3: Calculate the weight of oxalic acid
The molar mass of hydrated oxalic acid (H₂C₂O₄·2H₂O) is calculated as follows:
- Molar mass of H₂C₂O₄ = 2(1) + 2(12) + 4(16) = 90 g/mol
- Molar mass of 2H₂O = 2(2) + 2(16) = 36 g/mol
- Total molar mass of H₂C₂O₄·2H₂O = 90 + 36 = 126 g/mol

Using the formula:
\[ \text{Weight} = \text{Moles} \times \text{Molar Mass} \]
\[ \text{Weight} = 0.01 \, \text{moles} \times 126 \, \text{g/mol} = 1.26 \, \text{g} \]

### Final Answer
The weight of hydrated oxalic acid required for the complete neutralization of 100 ml of 0.2N NaOH solution is **1.26 grams**.

---