0.53 gm of `Na_(2)CO_(3)` has been dissolved in 100 ml of a sodium carbonate solution. The normality of the solution will be

Options

A. `(N)/(5)`
B. `(N)/(2)`
C. `(N)/(10)`
D. N

Explanation

To find the normality of the sodium carbonate (Na₂CO₃) solution, we can follow these steps:

### Step 1: Calculate the Molar Mass of Na₂CO₃
The molar mass of Na₂CO₃ can be calculated as follows:
- Sodium (Na): 23 g/mol × 2 = 46 g/mol
- Carbon (C): 12 g/mol × 1 = 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol

Adding these together:
\[ \text{Molar mass of Na}_2\text{CO}_3 = 46 + 12 + 48 = 106 \text{ g/mol} \]

### Step 2: Calculate the Equivalent Weight of Na₂CO₃
The equivalent weight of a substance is calculated by dividing its molar mass by the number of equivalents. For Na₂CO₃, it can donate 2 moles of Na⁺ ions and 1 mole of CO₃²⁻ ions, thus it has 2 equivalents:
\[ \text{Equivalent weight of Na}_2\text{CO}_3 = \frac{\text{Molar mass}}{\text{Number of equivalents}} = \frac{106 \text{ g/mol}}{2} = 53 \text{ g/equiv} \]

### Step 3: Use the Normality Formula
The formula for normality (N) is given by:
\[ N = \frac{\text{Weight of solute (g)} \times 1000}{\text{Equivalent weight (g/equiv)} \times \text{Volume of solution (L)}} \]
Here, the weight of Na₂CO₃ = 0.53 g, equivalent weight = 53 g/equiv, and the volume of solution = 100 mL = 0.1 L.

### Step 4: Substitute the Values into the Formula
Substituting the values into the normality formula:
\[ N = \frac{0.53 \, \text{g} \times 1000}{53 \, \text{g/equiv} \times 0.1 \, \text{L}} \]

### Step 5: Calculate Normality
Now, calculate the normality:
\[ N = \frac{0.53 \times 1000}{53 \times 0.1} = \frac{530}{5.3} = 100 \text{ N} \]

### Final Answer
The normality of the sodium carbonate solution is **10 N**.

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