Calculate the mass of sodium carbonate required to prepare 500 ml of a semi- normal solution

Options

A. `13.25 g`
B. `26.5 g`
C. `53 g`
D. `6.125 g`

Answer

To calculate the mass of sodium carbonate required to prepare 500 ml of a semi-normal solution, we can follow these steps:

### Step 1: Understand Normality
Normality (N) is defined as the number of equivalents of solute per liter of solution. A semi-normal solution is 0.5 N (half-normal).

### Step 2: Calculate the Equivalent Weight
The equivalent weight of a substance can be calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n \text{ factor}} \]

For sodium carbonate (Na2CO3):
- Molar mass of Na2CO3 = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 g/mol
- The n factor for sodium carbonate is determined by the total positive charge, which is 2 (from 2 sodium ions).

Thus, the equivalent weight of sodium carbonate is:
\[ \text{Equivalent Weight} = \frac{106 \text{ g/mol}}{2} = 53 \text{ g/equiv} \]

### Step 3: Calculate the Number of Equivalents Required
To find the number of equivalents needed for a semi-normal solution (0.5 N) in 500 ml (0.5 L):
\[ \text{Number of Equivalents} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Number of Equivalents} = 0.5 \text{ N} \times 0.5 \text{ L} = 0.25 \text{ equivalents} \]

### Step 4: Calculate the Mass of Sodium Carbonate
Now, we can calculate the mass of sodium carbonate required using the number of equivalents and the equivalent weight:
\[ \text{Mass} = \text{Number of Equivalents} \times \text{Equivalent Weight} \]
\[ \text{Mass} = 0.25 \text{ equivalents} \times 53 \text{ g/equiv} = 13.25 \text{ g} \]

### Final Answer
The mass of sodium carbonate required to prepare 500 ml of a semi-normal solution is **13.25 grams**.

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