Which one of the following aqueous solutions would have the highest freezing point?

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

Answer: (C)

Verified step by step guidance

1. Understand that the freezing point of a solution is lowered compared to pure solvent due to the presence of solute particles, a phenomenon called freezing point depression.
2. Recall the freezing point depression formula: \(\Delta T_f = i \cdot K_f \cdot m\), 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.
3. Identify the van't Hoff factor \(i\) for each solute: for glucose (C6H12O6), which does not dissociate, \(i = 1\); for KBr and NaCl, which dissociate into two ions each, \(i = 2\); for CaCl2, which dissociates into three ions, \(i = 3\).
4. Since all solutions have the same molarity (and assuming similar molality for comparison), the solution with the smallest \(i\) will have the smallest freezing point depression, thus the highest freezing point.
5. Conclude that the 0.10 M glucose solution, with \(i = 1\), will have the highest freezing point because it produces the fewest particles in solution, causing the least freezing point depression.