Which of the following aqueous solutions will have 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 KBr in 1.0 kg H2O
• D. 0.10 mol CaCl2 in 1.0 kg H2O

Answer: (D)

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 or identify 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 \(i=2\); KBr dissociates into K\(^+\) and Br\(^-\), so \(i=2\); CaCl\(_2\) dissociates into Ca\(^{2+}\) and 2 Cl\(^-\) ions, so \(i=3\).
3. Since all solutions have the same molality (\$0.10\( mol per 1.0 kg water), the difference in freezing point depression comes from the van't Hoff factor \)i\(. The greater the \)i$, the more particles in solution, and the greater the freezing point depression.
4. Compare the products \(i \cdot m\) for each solution: glucose (\$1 \times 0.10\(), NaCl (\)2 \times 0.10\(), KBr (\)2 \times 0.10\(), and CaCl\)_2\( (\)3 \times 0.10\(). The solution with CaCl\)_2$ has the highest effective particle concentration.
5. Conclude that the solution with CaCl\(_2\) will have the lowest freezing point because it produces the greatest freezing point depression due to the highest number of dissolved particles.