Which of the following aqueous solutions is expected to have the lowest freezing point?

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

Answer: (A)

Verified step by step guidance

1. Understand that the freezing point depression of a solution depends on the concentration of solute particles in the solution, not just the concentration of the compound itself. 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 glucose (\(C_6H_{12}O_6\)), which does not dissociate, \(i = 1\). - For NaCl, which dissociates into Na\(^+\) and Cl\(^-\), \(i = 2\). - For KNO\(_3\), which dissociates into K\(^+\) and NO\(_3^-\), \(i = 2\). - For CaCl\(_2\), which dissociates into Ca\(^{2+}\) and 2 Cl\(^-\) ions, \(i = 3\).
3. Since all solutions have the same molarity (0.10 M), and assuming similar molality, compare the effective concentration of particles by multiplying the molarity by the van't Hoff factor \(i\) for each solution. This gives the total concentration of dissolved particles that affect freezing point.
4. Calculate the relative freezing point depression for each solution using \(\Delta T_f = i \cdot K_f \cdot m\). The solution with the highest product of \(i \cdot m\) will have the greatest freezing point depression, thus the lowest freezing point.
5. Conclude that the solution with the highest van't Hoff factor and concentration of particles (in this case, 0.10 M CaCl\(_2\) with \(i=3\)) will have the lowest freezing point among the given options.