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

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

Answer: (C)

Verified step by step guidance

1. Recall that the freezing point depression of a solution depends on the number of solute particles dissolved in the solvent, 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 \(\Delta T_f\) 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: glucose (\(C_6H_{12}O_6\)) is a molecular compound and does not dissociate, so \(i = 1\); NaCl dissociates into Na\(^+\) and Cl\(^-\) ions, so \(i = 2\); CaCl\(_2\) dissociates into Ca\(^{2+}\) and 2 Cl\(^-\) ions, so \(i = 3\); KBr dissociates into K\(^+\) and Br\(^-\) ions, so \(i = 2\).
3. Since all solutions have the same molality (\$0.10\( mol per 1.0 kg of water), the difference in freezing point depression depends on the van't Hoff factor \)i\(. The greater the \)i$, the more particles in solution, and the greater the freezing point depression.
4. Compare the values of \(i\) for each solute: glucose (1), NaCl (2), CaCl\(_2\) (3), and KBr (2). The solution with CaCl\(_2\) has the highest \(i\) value, meaning it produces the most particles in solution and thus the greatest freezing point depression.
5. Conclude that the 0.10 mol CaCl\(_2\) in 1.0 kg H\(_2\)O solution has the lowest freezing point because it causes the largest freezing point depression due to its higher number of dissolved particles.