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

• A. 0.10 mol NaCl in 1.0 kg H2O
• B. 0.10 mol K2SO4 in 1.0 kg H2O
• C. 0.10 mol C6H12O6 (glucose) in 1.0 kg H2O
• D. 0.10 mol CaCl2 in 1.0 kg H2O

Answer: (C)

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, which is a colligative property. The formula for boiling point elevation is \(\Delta T_b = i K_b m\), 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. Calculate the van't Hoff factor \(i\) for each solute: for NaCl, which dissociates into Na\(^+\) and Cl\(^-\), \(i = 2\); for K\(_2\)SO\(_4\), which dissociates into 2 K\(^+\) and 1 SO\(_4^{2-}\), \(i = 3\); for glucose (C\(_6\)H\(_{12}\)O\(_6\)), which does not dissociate, \(i = 1\); and for CaCl\(_2\), which dissociates into Ca\(^{2+}\) and 2 Cl\(^-\), \(i = 3\).
3. Since all solutions have the same molality (\$0.10\( mol solute per \)1.0\( kg water), and the solvent is the same (water, so \)K_b\( is constant), the boiling point elevation depends directly on the van't Hoff factor \)i$.
4. Compare the values of \(i\) for each solution: NaCl (\(i=2\)), K\(_2\)SO\(_4\) (\(i=3\)), glucose (\(i=1\)), and CaCl\(_2\) (\(i=3\)). The solution with the smallest \(i\) will have the smallest boiling point elevation, and thus the lowest boiling point increase.
5. Conclude that the glucose solution, with \(i=1\), has the lowest boiling point elevation and therefore the lowest boiling point among the given solutions.