Which of the following aqueous solutions would have the higher boiling point?

• A. 1.0 mol NaCl in 1 kg H2O
• B. 0.5 mol NaCl in 1 kg H2O
• C. 1.0 mol glucose (C6H12O6) in 1 kg H2O
• D. 0.5 mol glucose (C6H12O6) in 1 kg H2O

Answer: (A)

Verified step by step guidance

1. Recall that the boiling point elevation of a solution depends on the number of solute particles dissolved in the solvent, not just the amount of solute in moles. This is described by the formula for boiling point elevation: \(\Delta T_b = i \cdot K_b \cdot m\), 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. Identify the van't Hoff factor \(i\) for each solute: For NaCl, which dissociates into Na\(^+\) and Cl\(^-\) ions, \(i\) is approximately 2. For glucose (C\(_6\)H\(_{12}\)O\(_6\)), which does not dissociate, \(i\) is 1.
3. Calculate the effective molality for each solution by multiplying the molality by the van't Hoff factor \(i\). For example, for 1.0 mol NaCl in 1 kg H\(_2\)O, the effective molality is \$1.0 \times 2 = 2.0$ mol/kg particles.
4. Compare the effective molalities of all solutions: 1.0 mol NaCl (2.0 mol/kg particles), 0.5 mol NaCl (1.0 mol/kg particles), 1.0 mol glucose (1.0 mol/kg particles), and 0.5 mol glucose (0.5 mol/kg particles). The solution with the highest effective molality will have the highest boiling point elevation.
5. Conclude that the solution with 1.0 mol NaCl in 1 kg H\(_2\)O has the highest boiling point because it produces the greatest number of dissolved particles, leading to the largest boiling point elevation.