Which of the following aqueous solutions has the highest boiling point?

• A. 0.1 M KNO_3
• B. 0.1 M C_6H_{12}O_6 (glucose)
• C. 0.1 M CaCl_2
• D. 0.1 M NaCl

Answer: (C)

Verified step by step guidance

1. Understand that the boiling point elevation depends on the number of dissolved particles in solution, which is related to the van't Hoff factor (i). The formula for boiling point elevation is \(\Delta T_b = i K_b m\), where \(\Delta T_b\) is the boiling point elevation, \(K_b\) is the ebullioscopic constant of the solvent, and \(m\) is the molality of the solution.
2. Identify the van't Hoff factor (i) for each solute, which represents the number of particles the compound dissociates into in solution: - For \(KNO_3\), it dissociates into \(K^+\) and \(NO_3^-\), so \(i = 2\). - For glucose (\(C_6H_{12}O_6\)), it does not dissociate, so \(i = 1\). - For \(CaCl_2\), it dissociates into \(Ca^{2+}\) and 2 \(Cl^-\) ions, so \(i = 3\). - For \(NaCl\), it dissociates into \(Na^+\) and \(Cl^-\), so \(i = 2\).
3. Since all solutions have the same molarity (0.1 M) and assuming similar molality, the solution with the highest van't Hoff factor will have the greatest boiling point elevation.
4. Compare the van't Hoff factors: \(CaCl_2\) has the highest value of 3, meaning it produces the most particles in solution, leading to the highest boiling point elevation.
5. Conclude that the 0.1 M \(CaCl_2\) solution will have the highest boiling point among the given options due to its greater number of dissolved particles increasing the boiling point more than the others.