Which of the following aqueous solutions is predicted to have the highest boiling point?

• A. 0.15 m AlCl_3
• B. 0.30 m C_6H_{12}O_6 (glucose)
• C. 0.20 m NaCl
• D. 0.10 m CaCl_2

Answer: (A)

Verified step by step guidance

1. Identify that the boiling point elevation depends on the colligative property formula: \(\Delta T_b = i \cdot K_b \cdot m\), where \(\Delta T_b\) is the boiling point elevation, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_b\) is the ebullioscopic constant of the solvent (water in this case), and \(m\) is the molality of the solution.
2. Determine the van't Hoff factor \(i\) for each solute based on its dissociation in water: - For glucose (\(C_6H_{12}O_6\)), a non-electrolyte, \(i = 1\). - For NaCl, which dissociates into Na\(^+\) and Cl\(^-\), \(i = 2\). - For CaCl\(_2\), which dissociates into Ca\(^{2+}\) and 2 Cl\(^-\), \(i = 3\). - For AlCl\(_3\), which dissociates into Al\(^{3+}\) and 3 Cl\(^-\), \(i = 4\).
3. Calculate the effective molality for each solution by multiplying the molality \(m\) by the van't Hoff factor \(i\): \(m_{effective} = i \times m\).
4. Compare the effective molalities of all solutions to determine which has the highest value, since the solution with the highest \(m_{effective}\) will have the greatest boiling point elevation.
5. Conclude that the solution with the highest effective molality corresponds to the highest boiling point elevation and thus the highest boiling point.