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

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

Answer: (A)

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 described 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: - Glucose (\(C_6H_{12}O_6\)) is a molecular compound and does not dissociate, so \(i = 1\). - Sodium chloride (NaCl) dissociates into Na\(^+\) and Cl\(^-\) ions, so \(i = 2\). - Potassium sulfate (\(K_2SO_4\)) dissociates into 2 K\(^+\) ions and 1 SO\(_4^{2-}\) ion, so \(i = 3\). - Calcium chloride (\(CaCl_2\)) dissociates into 1 Ca\(^{2+}\) ion and 2 Cl\(^-\) ions, so \(i = 3\).
3. Since all solutions have the same molarity (0.10 M), and assuming similar molality for simplicity, the boiling point elevation depends mainly on the van't Hoff factor \(i\). Calculate the effective particle concentration by multiplying molarity by \(i\) for each solution.
4. Compare the effective particle concentrations: the solution with the lowest \(i\) value will have the smallest boiling point elevation, thus the lowest boiling point elevation means the lowest boiling point increase from pure water.
5. Conclude that the solution with glucose (\(i=1\)) has the lowest boiling point elevation and therefore the lowest boiling point among the given options.