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V. Colligative Properties

Learn about colligative properties, including freezing point depression, boiling point elevation, vapor-pressure reduction, and osmotic pressure. Explore the calculations involved in determining these properties and molar mass.

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V. Colligative Properties

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  1. Ch. 13 - Solutions V. Colligative Properties

  2. A. Definition • Colligative Property • property that depends on the concentration (# of moles) of solute particles, not their chemical nature (identity). • For example: NaCl and MgSO4

  3. B. Types • 1. Freezing point depression (lowering) • 2. Boiling point elevation (raising) • 3. Vapor-pressure reduction • 4. Osmotic pressure

  4. B. Types 1. Freezing Point Depression View Flash animation.

  5. B. Types • Freezing Point Depression (tf) • f.p. of a solution is lower than f.p. of the pure solvent • tf -difference between the freezing point of the solution and the freezing point of the pure solvent

  6. B. Types • Applications • salting icy roads melts ice • making ice cream- lowers temp. • Antifreeze (ethylene glycol) • cars (-64°C to 136°C)

  7. B. Types 2. Boiling Point Elevation Solute particles weaken IMF in the solvent.

  8. B. Types • 2. Boiling Point Elevation (tb) • b.p. of a solution is higher than b.p. of the pure solvent • tb - difference between the boiling point of the solution and the boiling point of the pure solvent

  9. C. Calculations t: change in temperature (°C) k: constant based on the solvent (°C·kg/mol) m: molality (m) n: # of particles tf(b) = kf(b) · m · n 0.51 oC/m Boiling point k 1.86oC/m freezing point k

  10. C. Calculations • # of Particles: • Nonelectrolytes (covalent) • remain intact when dissolved • C6H12O6 C6H12O6n=1 • Electrolytes (ionic) • dissociate into ions when dissolved • 2 or more particles NaCl Na+ + Cl-n=2 MgCl2 Mg+2 + 2Cl-n=3

  11. C. Calculations • At what temperature will a solution that is composed of 0.73 moles of glucose (C6H12O6) in 225 g of water boil? GIVEN: b.p. = ? tb = ? kb = 0.51°C/ m WORK: m = 0.73mol ÷ 0.225kg tb = (0.51°C/ m)(3.2 m)(1) tb = 1.6°C b.p. = 100. °C + 1.6°C b.p. = 101.6°C m = 3.2 m n = 1 tb = kb · m · n

  12. C. Calculations • Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. GIVEN: f.p. = ? tf = ? kf = 1.86°C/ m WORK: m = 0.48mol ÷ 0.100kg tf = (1.86°C/ m)(4.8 m)(2) tf = 18°C f.p. = 0.00°C - 18°C f.p. = -18°C m = 4.8 m n = 2 tf = kf · m · n

  13. D. Molar Mass determination • Any colligative property can be used to determine the molar mass of an unknown (Exp. #24) • STRATEGY: Calculate number of moles required to produce observed change in colligative property

  14. D. Molar Mass determination • Use: tb = kb · m · n to find # of moles • tb = kb · x moles/kgsolv · n • Molar Mass ( M )= mass solute/x moles

  15. D. Molar Mass determination • Or use this formula: mass(g)solu · Kb(f) tb(f) · kgsolv Molar Mass (M) =

  16. D. Molar Mass determination • A 10.0 g sample of an unknown compound that does not dissociate is dissolved in 100.0 g of water. The boiling point of the solution is elevated 0.433 0C above the normal boiling point. What is the molar mass of the unknown sample?

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