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