Which of the following solutions will exhibit highest boiling point?

Options

A. `0.01 M Na_(2)SO_(4)`
B. `0.01M KNO_(3)`
C. `0.015 M` urea
D. `0.015 M` glucose

Explanation

To determine which of the given solutions will exhibit the highest boiling point, we will use the formula for boiling point elevation:

\[ \Delta T_B = K_B \cdot i \cdot m \]

Where:
- \(\Delta T_B\) is the boiling point elevation.
- \(K_B\) is the ebullioscopic constant (a property of the solvent).
- \(i\) is the van 't Hoff factor (the number of particles the solute dissociates into).
- \(m\) is the molality of the solution.

### Step-by-Step Solution:

1. **Identify the van 't Hoff factor (i) for each solute:**
- For **Na₂SO₄**:
- It dissociates into 2 Na⁺ and 1 SO₄²⁻, which gives a total of 3 ions.
- Therefore, \(i = 3\).
- For **KNO₃**:
- It dissociates into K⁺ and NO₃⁻, which gives a total of 2 ions.
- Therefore, \(i = 2\).
- For **Urea**:
- Urea does not dissociate in solution, so \(i = 1\).
- For **Glucose**:
- Glucose also does not dissociate in solution, so \(i = 1\).

2. **Calculate the product \(i \cdot m\) for each solution:**
- For **Na₂SO₄**:
- Given \(m = 0.01\),
- \(i \cdot m = 3 \cdot 0.01 = 0.03\).
- For **KNO₃**:
- Given \(m = 0.01\),
- \(i \cdot m = 2 \cdot 0.01 = 0.02\).
- For **Urea**:
- Given \(m = 0.015\),
- \(i \cdot m = 1 \cdot 0.015 = 0.015\).
- For **Glucose**:
- Given \(m = 0.015\),
- \(i \cdot m = 1 \cdot 0.015 = 0.015\).

3. **Compare the values of \(i \cdot m\):**
- Na₂SO₄: \(0.03\)
- KNO₃: \(0.02\)
- Urea: \(0.015\)
- Glucose: \(0.015\)

4. **Determine which solution has the highest boiling point:**
- The highest value of \(i \cdot m\) is for Na₂SO₄, which is \(0.03\).
- Therefore, Na₂SO₄ will exhibit the highest boiling point.

### Final Answer:
**Na₂SO₄ will exhibit the highest boiling point.**