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

• A. 0.10 molal KNO_3
• B. 0.10 molal C_6H_{12}O_6 (glucose)
• C. 0.10 molal NaCl
• D. 0.10 molal CaCl_2

Answer: (D)

Verified step by step guidance

1. Recall that the freezing point depression of a solution depends on the molality (m) of the solute, the van't Hoff factor (i), and the freezing point depression constant (K_f) of the solvent. The formula is: \[\Delta T_f = i \times K_f \times m\]
2. Identify the van't Hoff factor (i) for each solute, which represents the number of particles the solute dissociates into in solution: - For KNO_3, which dissociates into K^+ and NO_3^-, \[i = 2\] - For glucose (C_6H_{12}O_6), a non-electrolyte, \[i = 1\] - For NaCl, which dissociates into Na^+ and Cl^-, \[i = 2\] - For CaCl_2, which dissociates into Ca^{2+} and 2 Cl^-, \[i = 3\]
3. Since all solutions have the same molality (0.10 molal) and the same solvent (water, so K_f is constant), the freezing point depression depends directly on the van't Hoff factor. Calculate the effective concentration of particles as \[i \times m\] for each solution.
4. Compare the values of \[i \times m\] for each solution. The solution with the highest value will have the greatest freezing point depression and thus the lowest freezing point.
5. Conclude that 0.10 molal CaCl_2, with the highest van't Hoff factor (i = 3), produces the greatest freezing point depression and therefore has the lowest freezing point among the given solutions.