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

• A. 0.10 M KBr
• B. 0.10 M CaCl2
• C. 0.10 M C6H12O6 (glucose)
• D. 0.10 M NaCl

Answer: (B)

Verified step by step guidance

1. Recall that the freezing point depression of a solution depends on the number of solute particles in 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 KBr, which dissociates into K\(^+\) and Br\(^-\), \(i = 2\). - For CaCl\(_2\), which dissociates into Ca\(^{2+}\) and 2 Cl\(^-\) ions, \(i = 3\). - For glucose (C\(_6\)H\(_{12}\)O\(_6\)), a molecular compound that does not dissociate, \(i = 1\). - For NaCl, which dissociates into Na\(^+\) and Cl\(^-\), \(i = 2\).
3. Since all solutions have the same molarity (0.10 M), and assuming similar molality, the freezing point depression will be proportional to the van't Hoff factor \(i\). Therefore, the solution with the highest \(i\) will have the greatest freezing point depression and thus the lowest freezing point.
4. Compare the \(i\) values: CaCl\(_2\) has \(i=3\), which is greater than KBr and NaCl (\(i=2\)) and glucose (\(i=1\)). This means 0.10 M CaCl\(_2\) produces the most particles in solution and will cause the largest freezing point depression.
5. Conclude that among the given options, the 0.10 M CaCl\(_2\) solution will have the lowest freezing point due to its higher number of dissolved particles affecting the colligative property.