Which of the following aqueous solutions will have the lowest freezing point?

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

Answer: (B)

Verified step by step guidance

1. Understand that the freezing point depression depends on the number of solute particles in solution, not just the amount of substance. This is described by the formula for freezing point depression: \(\Delta T_f = i \cdot K_f \cdot m\), where \(i\) is the van't Hoff factor, \(K_f\) 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, which represents the number of particles the compound dissociates into in solution: for NaCl, \(i = 2\) (Na\(^+\) and Cl\(^-\)); for CaCl\(_2\), \(i = 3\) (Ca\(^{2+}\) and 2 Cl\(^-\)); for glucose (C\(_6\)H\(_{12}\)O\(_6\)), a nonelectrolyte, \(i = 1\); for KBr, \(i = 2\) (K\(^+\) and Br\(^-\)).
3. Since all solutions have the same molality (0.10 mol per 1.0 kg H\(_2\)O), compare the products \(i \cdot m\) for each to determine which has the greatest effect on freezing point depression.
4. Recognize that the solution with the highest \(i \cdot m\) value will have the greatest freezing point depression, thus the lowest freezing point.
5. Conclude that because CaCl\(_2\) has the highest van't Hoff factor (3) among the given solutes, its solution will have the lowest freezing point.