Which of the following aqueous solution should have the highest boiling point ?

Options

A. 1.0 M NaOH
B. `1.0 M Na_(2)SO_(4)`
C. `1.0 M NH_(4)NO_(3)`
D. `1.0 M KNO_(3)`

Explanation

To determine which of the given aqueous solutions has the highest boiling point, we need to analyze the dissociation of each compound in solution and apply the concept of colligative properties, specifically the elevation of boiling point.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Concentrations**:
We have four solutions:
- 1.0 M NaOH (sodium hydroxide)
- 1.0 M Na2SO4 (sodium sulfate)
- 1.0 M NH4NO3 (ammonium nitrate)
- 1.0 M KNO3 (potassium nitrate)

2. **Understand Colligative Properties**:
Colligative properties depend on the number of solute particles in a solution. The boiling point elevation is directly related to the number of particles produced when a solute dissolves in a solvent.

3. **Determine the Dissociation of Each Compound**:
- **NaOH**: Dissociates into 2 ions:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
(Van't Hoff factor, \(i = 2\))

- **Na2SO4**: Dissociates into 3 ions:
\[ \text{Na2SO4} \rightarrow 2 \text{Na}^+ + \text{SO}_4^{2-} \]
(Van't Hoff factor, \(i = 3\))

- **NH4NO3**: Dissociates into 2 ions:
\[ \text{NH4NO3} \rightarrow \text{NH}_4^+ + \text{NO}_3^- \]
(Van't Hoff factor, \(i = 2\))

- **KNO3**: Dissociates into 2 ions:
\[ \text{KNO3} \rightarrow \text{K}^+ + \text{NO}_3^- \]
(Van't Hoff factor, \(i = 2\))

4. **Calculate the Effective Number of Particles**:
Since all solutions are 1.0 M, we can simply look at the Van't Hoff factors:
- NaOH: \(i = 2\)
- Na2SO4: \(i = 3\)
- NH4NO3: \(i = 2\)
- KNO3: \(i = 2\)

5. **Determine the Highest Boiling Point**:
The boiling point elevation is highest for the solution with the greatest number of particles. Here, Na2SO4 produces the most particles (3 ions), thus it will have the highest boiling point.

### Conclusion:
The aqueous solution that should have the highest boiling point is **1.0 M Na2SO4**.