Which of the following aqueous solutions has the lowest freezing point?

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

Answer: (C)

Verified step by step guidance

1. Understand that the freezing point depression depends on the number of dissolved particles in solution, which is given by the formula for freezing point depression: \(\Delta T_f = i \cdot K_f \cdot m\), where \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_f\) is the freezing point depression constant of the solvent, and \(m\) is the molality of the solution.
2. Calculate the van't Hoff factor (\(i\)) for each solute: 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 \(i = 3\); glucose (C\(_6\)H\(_{12}\)O\(_6\)) does not dissociate, so \(i = 1\).
3. Since all solutions have the same molality (\(m = 0.10\) mol/kg) and the solvent is the same (water, so \(K_f\) is constant), compare the products \(i \cdot m\) for each solution to determine which has the greatest freezing point depression.
4. Recognize that the solution with the highest \(i \cdot m\) value will have the lowest freezing point because it produces the most particles in solution, causing the greatest freezing point depression.
5. Conclude that 0.10 mol CaCl\(_2\) in 1.0 kg H\(_2\)O has the lowest freezing point because it has the highest van't Hoff factor (\(i = 3\)), leading to the greatest freezing point depression.