Which of the following aqueous solutions has the lowest vapor pressure at 25°C?

• A. 1.0 mol glucose (C6H12O6) dissolved in 1 kg H2O
• B. 0.5 mol NaCl dissolved in 1 kg H2O
• C. 1.0 mol NaCl dissolved in 1 kg H2O
• D. 0.5 mol glucose (C6H12O6) dissolved in 1 kg H2O

Answer: (C)

Verified step by step guidance

1. Recall that the vapor pressure of a solution is lowered compared to pure solvent due to the presence of solute particles, a phenomenon known as vapor pressure lowering, which is a colligative property depending on the number of solute particles, not their identity.
2. Calculate the molality (moles of solute per kg of solvent) for each solution. Since the solvent mass is 1 kg in all cases, the molality equals the moles of solute for each solution.
3. Determine the van't Hoff factor (i) for each solute, which represents the number of particles the solute dissociates into in solution: glucose (C6H12O6) is a non-electrolyte with i = 1, while NaCl dissociates into Na⁺ and Cl⁻ ions with i ≈ 2.
4. Calculate the effective concentration of particles in solution by multiplying molality by the van't Hoff factor (i × molality) for each case. This value correlates with the extent of vapor pressure lowering.
5. Compare the values of i × molality for all solutions; the solution with the highest value will have the lowest vapor pressure because more solute particles cause greater vapor pressure lowering.