Given the following aqueous solutions at the same temperature, rank them from lowest to highest vapor pressure:A. 0.5 mol NaCl in 1.0 kg H2OB. 0.5 mol glucose (C6H12O6) in 1.0 kg H2OC. Pure H2OWhich sequence correctly ranks the solutions from lowest to highest vapor pressure?

• A. B, A, C
• B. A, B, C
• C. C, B, A
• D. A, C, B

Answer: (B)

Verified step by step guidance

1. Recall that vapor pressure lowering is a colligative property, which depends on the number of solute particles dissolved in the solvent, not their identity.
2. Calculate the molality of each solution, which is moles of solute per kilogram of solvent. Both solutions A and B have 0.5 mol solute in 1.0 kg water, so their molality is 0.5 mol/kg.
3. Consider the nature of the solutes: NaCl is an ionic compound that dissociates into ions in solution, while glucose is a molecular compound that does not dissociate.
4. Determine the total number of solute particles in each solution: NaCl dissociates into 2 ions (Na⁺ and Cl⁻), so 0.5 mol NaCl produces 1.0 mol particles; glucose remains as 0.5 mol particles.
5. Use Raoult's law, which states that vapor pressure lowering is proportional to the mole fraction of solute particles. More particles mean greater lowering, so rank the solutions from lowest to highest vapor pressure as: A (most particles), B (fewer particles), C (pure solvent, no solute).