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

• A. 0.5 mol glucose (C6H12O6) in 1 kg H2O
• B. 0.5 mol CaCl2 in 1 kg H2O
• C. 0.5 mol KBr in 1 kg H2O
• D. 0.5 mol NaCl in 1 kg H2O

Answer: (B)

Verified step by step guidance

1. Understand that the boiling point elevation depends on the number of solute particles in solution, not just the amount of substance. This is a colligative property, which depends on the total concentration of dissolved particles.
2. Recall the boiling point elevation formula: \(\Delta T_b = i \cdot K_b \cdot m\), where \(\Delta T_b\) is the boiling point elevation, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_b\) is the boiling point elevation constant of the solvent, and \(m\) is the molality of the solution.
3. Calculate or identify the van't Hoff factor \(i\) for each solute: glucose does not dissociate, so \(i=1\); NaCl dissociates into Na\(^+\) and Cl\(^-\), so \(i=2\); KBr dissociates into K\(^+\) and Br\(^-\), so \(i=2\); CaCl\(_2\) dissociates into Ca\(^{2+}\) and 2 Cl\(^-\), so \(i=3\).
4. Since all solutions have the same molality (0.5 mol per 1 kg H\(_2\)O), compare the products \(i \cdot m\) for each: glucose (0.5), NaCl (1.0), KBr (1.0), CaCl\(_2\) (1.5). The solution with the highest \(i \cdot m\) will have the highest boiling point elevation.
5. Conclude that the 0.5 mol CaCl\(_2\) solution has the highest boiling point because it produces the greatest number of dissolved particles, leading to the largest boiling point elevation.