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

• A. 0.10 mol CaCl2 in 1.0 kg H2O
• B. 0.20 mol glucose (C6H12O6) in 1.0 kg H2O
• C. 0.10 mol AlCl3 in 1.0 kg H2O
• D. 0.20 mol NaCl in 1.0 kg H2O

Answer: (C)

Verified step by step guidance

1. Identify that the problem involves boiling point elevation, which is a colligative property depending on the number of solute particles in solution, not their identity.
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 ebullioscopic constant of the solvent, and \(m\) is the molality of the solution.
3. Calculate the molality (\(m\)) for each solution, which is moles of solute per kilogram of solvent. Here, all solutions have 1.0 kg of water, so molality equals the moles of solute given.
4. Determine the van't Hoff factor (\(i\)) for each solute based on its dissociation in water: - CaCl\(_2\) dissociates into 3 ions (\(Ca^{2+}\) and 2 \(Cl^-\)), so \(i=3\). - Glucose does not dissociate, so \(i=1\). - AlCl\(_3\) dissociates into 4 ions (\(Al^{3+}\) and 3 \(Cl^-\)), so \(i=4\). - NaCl dissociates into 2 ions (\(Na^+\) and \(Cl^-\)), so \(i=2\).
5. Calculate the product \(i \times m\) for each solution to compare their boiling point elevations. The solution with the highest \(i \times m\) value will have the highest boiling point.