Which of the following solutions has the higher boiling point?

• A. 1.0 mol NaCl dissolved in 1.0 kg water
• B. 1.0 mol glucose (C6H12O6) dissolved in 1.0 kg water
• C. Pure water

Answer: (A)

Verified step by step guidance

1. Identify that the boiling point elevation depends on the colligative properties of the solution, which are influenced by the number of solute particles in the solution, not their identity.
2. Recall the boiling point elevation 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 boiling point elevation constant of the solvent, and \(m\) is the molality of the solution.
3. Determine the van't Hoff factor \(i\) for each solute: for NaCl, which dissociates into Na\(^+\) and Cl\(^-\) ions, \(i\) is approximately 2; for glucose, which does not dissociate, \(i\) is 1; for pure water, there are no solute particles, so \(i\) is 0.
4. Calculate the molality \(m\) for each solution, which is moles of solute per kilogram of solvent. Both solutions have 1.0 mol solute in 1.0 kg water, so \(m = 1.0\) mol/kg for both.
5. Compare the boiling point elevations by plugging values into the formula: since \(K_b\) and \(m\) are the same for both solutions, the solution with the higher \(i\) (NaCl) will have the higher boiling point elevation, and thus the highest boiling point.