Which of the following solutions has highest bolling point?

Options

A. 0.1 m glucose in water
B. 0.1 m sucrose in water
C. 0.1 m NaCl in water
D. `0.1 m CaCl_2 in water`

Answer

To determine which of the following solutions has the highest boiling point, we need to analyze the given options based on the concept of boiling point elevation, which is influenced by the number of solute particles in the solution.

### Step-by-Step Solution:

1. **Understanding Boiling Point Elevation**:
The elevation in boiling point (ΔTb) is given by the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(K_b\) = ebullioscopic constant of the solvent (water in this case)
- \(m\) = molality of the solution

2. **Identifying the Solutions**:
The options provided are:
- 0.1 molal glucose in water
- 0.1 molal sucrose in water
- 0.1 molal NaCl in water
- 0.1 molal CaCl2 in water

3. **Calculating the van 't Hoff Factor (i)**:
- **Glucose (C6H12O6)**: Does not dissociate in solution, so \(i = 1\).
- **Sucrose (C12H22O11)**: Also does not dissociate, so \(i = 1\).
- **NaCl**: Dissociates into Na⁺ and Cl⁻, so \(i = 2\).
- **CaCl2**: Dissociates into Ca²⁺ and 2 Cl⁻, so \(i = 3\).

4. **Calculating ΔTb for Each Solution**:
- For glucose:
\[ \Delta T_b = 1 \cdot K_b \cdot 0.1 \]
- For sucrose:
\[ \Delta T_b = 1 \cdot K_b \cdot 0.1 \]
- For NaCl:
\[ \Delta T_b = 2 \cdot K_b \cdot 0.1 \]
- For CaCl2:
\[ \Delta T_b = 3 \cdot K_b \cdot 0.1 \]

5. **Comparing the Elevation in Boiling Points**:
- Glucose and sucrose both have the same ΔTb: \(0.1 K_b\).
- NaCl has a ΔTb of \(0.2 K_b\).
- CaCl2 has the highest ΔTb of \(0.3 K_b\).

6. **Conclusion**:
Since the boiling point elevation (ΔTb) is highest for the CaCl2 solution, it will also have the highest boiling point. Therefore, the solution with the highest boiling point is:
\[ \text{0.1 molal CaCl2 in water} \]

### Final Answer:
The solution with the highest boiling point is **0.1 molal CaCl2 in water**.

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