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

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

Answer: (B)

Verified step by step guidance

1. Recall that the freezing point depression of a solution depends on the number of solute particles dissolved in the solvent, not just the concentration of the solute. This is a colligative property described by the formula \(\Delta T_f = i K_f m\), 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 nonelectrolytes like glucose (\(\mathrm{C_6H_{12}O_6}\)), \(i = 1\) because it does not dissociate; for electrolytes like \(\mathrm{NaCl}\) and \(\mathrm{KBr}\), which dissociate into two ions, \(i = 2\); for \(\mathrm{CaCl_2}\), which dissociates into three ions (\(\mathrm{Ca^{2+}}\) and 2 \(\mathrm{Cl^-}\)), \(i = 3\).
3. Since all solutions have the same molarity (0.10 M), and assuming similar molality, compare the effective concentration of particles by multiplying molarity by the van't Hoff factor: \(\text{effective particles} = i \times 0.10\) M.
4. The solution with the highest number of dissolved particles will have the greatest freezing point depression (lowest freezing point), and the one with the fewest particles will have the smallest freezing point depression (highest freezing point).
5. Therefore, glucose solution, which does not dissociate and has the smallest \(i\), will have the highest freezing point among the given options.