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Ch.13 - Solutions & Their Properties
All textbooksMcMurry 8th EditionCh.13 - Solutions & Their PropertiesProblem 91
Chapter 13, Problem 91

Which of the following aqueous solutions has the (a) higher freezing point, (b) higher boiling point, (c) lower vapor pressure: 0.50 m sucrose (C12H22O11) or 0.35 m HNO3?

Verified step by step guidance
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Step 1: Understand the colligative properties involved. Freezing point depression, boiling point elevation, and vapor pressure lowering are colligative properties that depend on the number of solute particles in a solution.
Step 2: Determine the van't Hoff factor (i) for each solute. Sucrose (C12H22O11) is a non-electrolyte, so it does not dissociate in solution, giving it a van't Hoff factor of i = 1. HNO3 is a strong acid and dissociates completely in water to form H+ and NO3-, giving it a van't Hoff factor of i = 2.
Step 3: Calculate the effective molality for each solution. For sucrose, the effective molality is 0.50 m * 1 = 0.50 m. For HNO3, the effective molality is 0.35 m * 2 = 0.70 m.
Step 4: Compare the colligative properties. (a) Higher freezing point: The solution with the lower effective molality will have the higher freezing point. (b) Higher boiling point: The solution with the higher effective molality will have the higher boiling point. (c) Lower vapor pressure: The solution with the higher effective molality will have the lower vapor pressure.
Step 5: Conclude based on the effective molalities. Use the calculated effective molalities to determine which solution has the higher freezing point, higher boiling point, and lower vapor pressure.
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