Which of the following aqueous solutions has the highest boiling point?

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

Answer: (D)

Verified step by step guidance

1. Understand that the boiling point elevation depends on the number of solute particles in solution, not their identity. This is a colligative property described by the 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.
2. Calculate or identify the van't Hoff factor (\(i\)) for each solute: for NaCl, which dissociates into Na\(^+\) and Cl\(^-\), \(i = 2\); for glucose, which does not dissociate, \(i = 1\); for CaCl\(_2\), which dissociates into Ca\(^{2+}\) and 2 Cl\(^-\) ions, \(i = 3\); and for pure water, \(i = 0\) since there is no solute.
3. Note that the molality (\(m\)) is the same for all solutions (0.5 mol solute per 1 kg water), so the difference in boiling point elevation depends mainly on the van't Hoff factor \(i\).
4. Recognize that the boiling point elevation constant \(K_b\) is the same for all solutions since the solvent is water in each case.
5. Conclude that the solution with the highest product of \(i \cdot m\) will have the highest boiling point elevation, and therefore the highest boiling point. Since CaCl\(_2\) has the largest \(i\) value (3) at the same molality, it will cause the greatest boiling point elevation.