The mass of oxalic acid crystals `(H_(2)C_(2)O_(4).2H_(2)O)` required to prepare 50 mL of a 0.2 N solution is:

Options

A. 4.5 g
B. 6.3 g
C. 0.63 g
D. 0.45 g

Explanation

To determine the mass of oxalic acid crystals `(H2C2O4.2H2O)` required to prepare 50 mL of a 0.2 N solution, we can follow these steps:

### Step 1: Understand Normality
Normality (N) is defined as the number of gram equivalents of solute per liter of solution.

### Step 2: Calculate the Gram Equivalent of Oxalic Acid
The formula for calculating the gram equivalent is:
\[ \text{Gram Equivalent} = \frac{\text{Molecular Weight}}{\text{Basicity}} \]

- **Molecular Weight of Oxalic Acid Crystals (H2C2O4.2H2O)**:
- H: 1 g/mol × 2 = 2 g/mol
- C: 12 g/mol × 2 = 24 g/mol
- O: 16 g/mol × 4 = 64 g/mol
- H2O: 18 g/mol × 2 = 36 g/mol
- Total = 2 + 24 + 64 + 36 = 126 g/mol

- **Basicity**: Oxalic acid has 2 replaceable hydrogen ions, so its basicity = 2.

Using these values:
\[ \text{Gram Equivalent} = \frac{126 \text{ g/mol}}{2} = 63 \text{ g/equiv} \]

### Step 3: Calculate the Mass for 1 Liter of 0.2 N Solution
For a 1 N solution, we need 63 grams of oxalic acid. For a 0.2 N solution, we can find the required mass (x) as follows:
\[ x = 63 \text{ g} \times 0.2 \text{ N} = 12.6 \text{ g} \]

### Step 4: Calculate the Mass for 50 mL of 0.2 N Solution
Now, since we need only 50 mL of this solution, we can use a proportion:
\[ \text{Mass required for 50 mL} = \frac{50 \text{ mL}}{1000 \text{ mL}} \times 12.6 \text{ g} \]

Calculating this gives:
\[ \text{Mass required} = 0.05 \times 12.6 \text{ g} = 0.63 \text{ g} \]

### Conclusion
The mass of oxalic acid crystals `(H2C2O4.2H2O)` required to prepare 50 mL of a 0.2 N solution is **0.63 grams**.

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