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

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

Answer: (A)

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 a colligative property, which depends on the total concentration of dissolved particles.
2. Recall 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.
3. Determine 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 \(i=3\).
4. Since all solutions have the same molality (\$0.10\( mol per \)1.0\( kg \)H_2O\(), compare the products \)i \cdot m$ for each to find which has the greatest value, as this will cause the greatest freezing point depression.
5. Identify the solution with the highest \(i \cdot m\) value as the one with the lowest freezing point. In this case, \(CaCl_2\) with \(i=3\) and \(m=0.10\) will have the largest freezing point depression.