Calculate the normally of 10.6 % (w/v) `Na_2CO_3` in aqueous solutions.

Answer

To calculate the normality of a 10.6% (w/v) solution of sodium carbonate (Na₂CO₃), we can follow these steps:

### Step 1: Calculate the Molecular Mass of Sodium Carbonate (Na₂CO₃)
- Sodium (Na): 23 g/mol (2 atoms) = 2 × 23 = 46 g/mol
- Carbon (C): 12 g/mol (1 atom) = 12 g/mol
- Oxygen (O): 16 g/mol (3 atoms) = 3 × 16 = 48 g/mol

**Total Molecular Mass = 46 + 12 + 48 = 106 g/mol**

### Step 2: Determine the Equivalent Mass of Sodium Carbonate
- The equivalent mass is calculated using the formula:

\[ \text{Equivalent Mass} = \frac{\text{Molar Mass}}{n} \]

- For sodium carbonate, it can donate 2 moles of Na⁺ ions (since there are 2 sodium ions). Thus, the n-factor (number of equivalents) is 2.

\[ \text{Equivalent Mass} = \frac{106 \text{ g/mol}}{2} = 53 \text{ g/equiv} \]

### Step 3: Calculate the Number of Equivalents in the Solution
- Given that the solution is 10.6% (w/v), this means there are 10.6 grams of Na₂CO₃ in 100 mL of solution.

- The number of equivalents can be calculated using the formula:

\[ \text{Number of Equivalents} = \frac{\text{Given Mass}}{\text{Equivalent Mass}} = \frac{10.6 \text{ g}}{53 \text{ g/equiv}} \approx 0.2 \text{ equivalents} \]

### Step 4: Calculate the Normality of the Solution
- Normality (N) is defined as the number of equivalents of solute per liter of solution. Since we have 0.2 equivalents in 100 mL (0.1 L), we can calculate normality as follows:

\[ \text{Normality} = \frac{\text{Number of Equivalents}}{\text{Volume of Solution in Liters}} = \frac{0.2 \text{ equivalents}}{0.1 \text{ L}} = 2 \text{ N} \]

### Final Answer
The normality of the 10.6% (w/v) sodium carbonate solution is **2 N**.

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