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

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

Answer: (A)

Verified step by step guidance

1. Recall that the freezing point depression of a solution depends on the molality of the solute and the number of particles the solute dissociates into. The formula for freezing point depression is \(\Delta T_f = i \cdot K_f \cdot m\), 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 glucose (\(\mathrm{C_6H_{12}O_6}\)), it does not dissociate, so \(i = 1\). - For \(\mathrm{NaCl}\), it dissociates into 2 ions (\(\mathrm{Na^+}\) and \(\mathrm{Cl^-}\)), so \(i = 2\).
3. Since all solutions have the same molality (\$0.10\( molal) and are in the same solvent (water), the freezing point depression depends directly on the van't Hoff factor \)i\(. Calculate the effective concentration of particles as \)i \times m$ for each solution.
4. Compare the values of \(i \times m\) for each solution to determine which has the greatest freezing point depression (largest \(\Delta T_f\)), which corresponds to the lowest freezing point.
5. Conclude that the solution with the highest \(i \times m\) value will have the lowest freezing point. Therefore, the solution with \(\mathrm{CaCl_2}\), having \(i = 3\), will have the greatest freezing point depression and thus the lowest freezing point.