5 g sample contain only `Na_(2)CO_(3)` and `Na_(2)SO_(4)` . This sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralizes 20 mL of 0.1 M `H_(2)SO_(4)`.
Calcalute the % of `Na_(2)SO_(4)` in the sample .

Options

A. 42.4
B. 57.6
C. 36.2
D. None of these

Answer

To solve the problem, we need to calculate the percentage of sodium sulfate (Na₂SO₄) in a 5 g sample that contains only Na₂CO₃ and Na₂SO₄. We will use the information given about the neutralization reaction with sulfuric acid (H₂SO₄) to find the required percentage.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between sodium carbonate (Na₂CO₃) and sulfuric acid (H₂SO₄) is:
\[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{CO}_3 \]
Sodium sulfate (Na₂SO₄) does not react with H₂SO₄.

2. **Calculate Milliequivalents of H₂SO₄**:
We know that 25 mL of the solution neutralizes 20 mL of 0.1 M H₂SO₄. First, we calculate the milliequivalents of H₂SO₄ used:
\[ \text{Milliequivalents of H}_2\text{SO}_4 = \text{Volume (mL)} \times \text{Molarity (M)} \times \text{N factor} \]
The N factor for H₂SO₄ is 2 (since it can donate 2 H⁺ ions):
\[ \text{Milliequivalents of H}_2\text{SO}_4 = 20 \, \text{mL} \times 0.1 \, \text{M} \times 2 = 4 \, \text{meq} \]

3. **Milliequivalents of Na₂CO₃**:
Since Na₂CO₃ reacts with H₂SO₄ in a 1:1 ratio, the milliequivalents of Na₂CO₃ will also be 4 meq.

4. **Calculate Millimoles of Na₂CO₃**:
To find the millimoles of Na₂CO₃, we use the relation:
\[ \text{Millimoles of Na}_2\text{CO}_3 = \frac{\text{Milliequivalents}}{\text{N factor}} \]
The N factor for Na₂CO₃ is also 2:
\[ \text{Millimoles of Na}_2\text{CO}_3 = \frac{4 \, \text{meq}}{2} = 2 \, \text{mmol} \]

5. **Calculate Total Millimoles in 250 mL**:
Since 25 mL of the solution contains 2 mmol of Na₂CO₃, we can find the total millimoles in 250 mL:
\[ \text{Total millimoles in 250 mL} = 2 \, \text{mmol} \times \frac{250 \, \text{mL}}{25 \, \text{mL}} = 20 \, \text{mmol} \]

6. **Calculate the Mass of Na₂CO₃**:
The molar mass of Na₂CO₃ is approximately 106 g/mol. Therefore, the mass of Na₂CO₃ in the sample is:
\[ \text{Mass of Na}_2\text{CO}_3 = \text{Millimoles} \times \text{Molar mass} = 20 \, \text{mmol} \times 106 \, \text{g/mol} = 2.12 \, \text{g} \]

7. **Calculate the Mass of Na₂SO₄**:
The total mass of the sample is 5 g. Thus, the mass of Na₂SO₄ can be calculated as:
\[ \text{Mass of Na}_2\text{SO}_4 = \text{Total mass} - \text{Mass of Na}_2\text{CO}_3 = 5 \, \text{g} - 2.12 \, \text{g} = 2.88 \, \text{g} \]

8. **Calculate the Percentage of Na₂SO₄**:
Finally, we can find the percentage of Na₂SO₄ in the sample:
\[ \text{Percentage of Na}_2\text{SO}_4 = \left( \frac{\text{Mass of Na}_2\text{SO}_4}{\text{Total mass}} \right) \times 100 = \left( \frac{2.88 \, \text{g}}{5 \, \text{g}} \right) \times 100 = 57.6\% \]

### Final Answer:
The percentage of Na₂SO₄ in the sample is **57.6%**.