Which of the following aqueous solutions should have the lowest boiling point?

• A. 0.10 M K2SO4
• B. 0.10 M NaCl
• C. 0.10 M CaCl2
• D. 0.10 M C6H12O6 (glucose)

Answer: (D)

Verified step by step guidance

1. Recall that the boiling point elevation 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 and is given by the formula: \(\Delta T_b = i K_b m\), 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 boiling point elevation constant of the solvent, and \(m\) is the molality of the solution.
2. Identify the van't Hoff factor (\(i\)) for each solute: - For \(\mathrm{K_2SO_4}\), it dissociates into 2 K\(^+\) and 1 SO\(_4^{2-}\) ions, so \(i = 3\). - For NaCl, it dissociates into Na\(^+\) and Cl\(^-\), so \(i = 2\). - For CaCl\(_2\), it dissociates into Ca\(^{2+}\) and 2 Cl\(^-\) ions, so \(i = 3\). - For glucose (C\(_6\)H\(_{12}\)O\(_6\)), it does not dissociate, so \(i = 1\).
3. Since all solutions have the same molarity (0.10 M), and assuming similar molality, the key difference in boiling point elevation comes from the van't Hoff factor \(i\). The higher the \(i\), the more particles in solution, and the greater the boiling point elevation.
4. Compare the \(i\) values: K\(_2\)SO\(_4\) and CaCl\(_2\) have \(i=3\), NaCl has \(i=2\), and glucose has \(i=1\). Therefore, glucose produces the fewest particles in solution and will cause the smallest increase in boiling point.
5. Because the question asks for the solution with the lowest boiling point, it corresponds to the solution with the smallest boiling point elevation, which is the one with glucose (C\(_6\)H\(_{12}\)O\(_6\)).