Which of the following aqueous solutions would have the lowest freezing point?

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

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 in solution. 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 (water in this case), 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 1 \(\mathrm{Ca^{2+}}\) and 2 \(\mathrm{Cl^-}\) ions, so \(i = 3\). - For \(\mathrm{NaCl}\), it dissociates into 1 \(\mathrm{Na^+}\) and 1 \(\mathrm{Cl^-}\) ion, so \(i = 2\). - For glucose (\(\mathrm{C_6H_{12}O_6}\)), it does not dissociate, so \(i = 1\). - For \(\mathrm{KNO_3}\), it dissociates into 1 \(\mathrm{K^+}\) and 1 \(\mathrm{NO_3^-}\) ion, so \(i = 2\).
3. Since all solutions have the same molality (\$0.10\( molal), compare the products \)i \cdot m\( for each solution to determine which will have the greatest freezing point depression. The solution with the highest \)i \cdot m$ will have the lowest freezing point.
4. Calculate the effective molality (or total particle concentration) for each solution by multiplying the molality by the van't Hoff factor: - \(\mathrm{CaCl_2}\): \$3 \times 0.10 = 0.30$ - \(\mathrm{NaCl}\): \$2 \times 0.10 = 0.20$ - Glucose: \$1 \times 0.10 = 0.10$ - \(\mathrm{KNO_3}\): \$2 \times 0.10 = 0.20$
5. Conclude that the solution with the highest effective molality (\$0.30\( for \)\mathrm{CaCl_2}$) will have the greatest freezing point depression and thus the lowest freezing point.