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, follow these steps:

### Step 1: Understand the Definition of Semi-Normal Solution
A semi-normal solution (0.5 N) means that the normality of the solution is half of 1 N.

### Step 2: Calculate the Equivalent Weight of Sodium Carbonate
The molecular weight of sodium carbonate (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 gives:
\[ \text{Molecular weight of Na2CO3} = 46 + 12 + 48 = 106 \, \text{g/mol} \]

The equivalent weight of sodium carbonate (Na2CO3) is calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n}} \]
For sodium carbonate, n = 2 (as it can donate 2 moles of Na+ ions).

Thus,
\[ \text{Equivalent weight} = \frac{106}{2} = 53 \, \text{g/equiv} \]

### Step 3: Use the Formula for Normality
The formula for normality (N) is given by:
\[ N = \frac{\text{weight (g)} \times 1000}{\text{equivalent weight (g/equiv)} \times \text{volume (ml)}} \]

For a semi-normal solution (0.5 N), we rearrange the formula to find the weight:
\[ \text{Weight (g)} = \frac{N \times \text{equivalent weight (g/equiv)} \times \text{volume (ml)}}{1000} \]

### Step 4: Substitute the Values
Substituting the values we have:
- N = 0.5
- Equivalent weight = 53 g/equiv
- Volume = 500 ml

Calculating the weight:
\[ \text{Weight (g)} = \frac{0.5 \times 53 \times 500}{1000} \]

### Step 5: Perform the Calculation
Calculating the above expression:
\[ \text{Weight (g)} = \frac{0.5 \times 53 \times 500}{1000} = \frac{13250}{1000} = 13.25 \, \text{g} \]

### 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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