Which among the following has highest boiling point ?

Options

A. 1 M glucose
B. 1 M KCl
C. 1 M Al`(NO_(3))_(3)`
D. 1 M `Na_(2)SO_(4)`

Answer

To determine which among the given options has the highest boiling point, we can use the formula for boiling point elevation, which is given by:

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

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

Since the concentration (1 molar) is the same for all options, and \(K_B\) is a constant for the solvent, the boiling point elevation \(\Delta T_B\) will be directly proportional to the van't Hoff factor \(i\).

### Step-by-Step Solution:

1. **Identify the van't Hoff factor for each solute:**
- **Glucose (C₆H₁₂O₆)**: Glucose does not dissociate in solution. Therefore, \(i = 1\).
- **KCl (Potassium Chloride)**: KCl dissociates into K⁺ and Cl⁻. Thus, \(i = 2\) (1 K⁺ + 1 Cl⁻).
- **Aluminium Nitrate (Al(NO₃)₃)**: Aluminium nitrate dissociates into Al³⁺ and 3 NO₃⁻ ions. Therefore, \(i = 4\) (1 Al³⁺ + 3 NO₃⁻).
- **Na₂SO₄ (Sodium Sulfate)**: Sodium sulfate dissociates into 2 Na⁺ and SO₄²⁻. Thus, \(i = 3\) (2 Na⁺ + 1 SO₄²⁻).

2. **Compare the van't Hoff factors:**
- Glucose: \(i = 1\)
- KCl: \(i = 2\)
- Aluminium Nitrate: \(i = 4\)
- Sodium Sulfate: \(i = 3\)

3. **Determine which has the highest \(i\):**
- The highest van't Hoff factor is for Aluminium Nitrate, which has \(i = 4\).

4. **Conclusion:**
- Since the boiling point elevation \(\Delta T_B\) is directly proportional to \(i\), the solution with the highest boiling point will be the one with the highest van't Hoff factor. Therefore, **1 molar Aluminium Nitrate** will have the highest boiling point among the given options.

### Final Answer:
**1 molar Aluminium Nitrate has the highest boiling point.**

---