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

• A. 0.1 M CaCl_2
• B. 0.1 M KNO_3
• C. 0.1 M NaCl
• D. 0.1 M C_6H_{12}O_6 (glucose)

Answer: (A)

Verified step by step guidance

1. Understand that the freezing point depression of a solution depends on the number of dissolved particles in the solution, not just the concentration of the compound. 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. Identify the van't Hoff factor (\(i\)) for each solute: - For \(\mathrm{CaCl_2}\), it dissociates into 3 ions (\(\mathrm{Ca^{2+}}\) and 2 \(\mathrm{Cl^-}\)), so \(i = 3\). - For \(\mathrm{KNO_3}\), it dissociates into 2 ions (\(\mathrm{K^+}\) and \(\mathrm{NO_3^-}\)), so \(i = 2\). - For \(\mathrm{NaCl}\), it dissociates into 2 ions (\(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\)), so \(i = 2\). - For glucose (\(\mathrm{C_6H_{12}O_6}\)), it does not dissociate, so \(i = 1\).
3. Since all solutions have the same molarity (0.1 M), and assuming similar molality, compare the effective concentration of particles by multiplying molarity by the van't Hoff factor: \(i \times 0.1\) M for each solution.
4. The solution with the highest effective particle concentration will have the greatest freezing point depression, thus the lowest freezing point.
5. Conclude that the solution with \(\mathrm{CaCl_2}\), having the highest van't Hoff factor of 3, produces the most particles in solution and therefore has the lowest freezing point among the options.