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IV. Colligative Properties of Solutions (p. 498 – 504)

Ch. 14 – Mixtures & Solutions. IV. Colligative Properties of Solutions (p. 498 – 504). A. Definition. Colligative Property property that depends on the concentration of solute particles, not their identity Examples: vapor pressure, freezing point, boiling point. B. Types. B. Types.

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IV. Colligative Properties of Solutions (p. 498 – 504)

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  1. Ch. 14 – Mixtures & Solutions IV. Colligative Properties of Solutions(p. 498 – 504)

  2. A. Definition • Colligative Property • property that depends on the concentration of solute particles, not their identity • Examples: vapor pressure, freezing point, boiling point

  3. B. Types

  4. B. Types • Freezing Point Depression (tf) • f.p. of a solution is lower than f.p. of the pure solvent • Boiling Point Elevation (tb) • b.p. of a solution is higher than b.p. of the pure solvent

  5. B. Types • Applications • salting icy roads • making ice cream • antifreeze • cars (-64°C to 136°C) • fish & insects

  6. C. Calculations t:change in temperature (°C) i: Van’t Hoff Factor (VHF), the number of particles into which the solute dissociates m: molality (m) K:constant based on the solvent (°C·kg/mol) or (°C/m) t = i · m · K

  7. C. Calculations • t • Change in temperature • Not actual freezing point or boiling point • Change from FP or BP of pure solvent • Freezing Point (FP) • tf is always subtracted from FP of pure solvent • Boiling Point (BP) • tb is always added to BP of pure solvent

  8. C. Calculations • i – VHF • Nonelectrolytes (covalent) • remain intact when dissolved • 1 particle • Electrolytes (ionic) • dissociate into ions when dissolved • 2 or more particles

  9. C. Calculations • i – VHF • Examples • CaCl2 • Ethanol C2H5OH • Al2(SO4)3 • Methane CH4 • i = • 3 • 1 • 5 • 1

  10. C. Calculations • K – molal constant • Kf – molal freezing point constant • Changes for every solvent • 1.86 °C·kg/mol (or °C/m) for water • Kb – molal boiling point constant • Changes for every solvent • 0.512 °C·kg/mol (or °C/m) for water

  11. C. Calculations: Recap! t = i · m · K • t : subtract from F.P. add to B.P. • i – VHF : covalent = 1 ionic > 2 • K : Kf water = 1.86 °C·kg/mol Kb water = 0.512 °C·kg/mol

  12. C. Calculations • At what temperature will a solution that is composed of 0.730 moles of glucose in 225 g of water boil? GIVEN: b.p. = ? tb = ? i = 1 WORK: m = 0.730 mol ÷ 0.225 kg 100 + tb tb = (1)(3.24m)(0.512°C/m) tb = 1.66°C b.p. = 100.00°C + 1.66°C b.p. = 101.66°C m = 3.24m Kb = 0.512°C/m tb = i · m · Kb

  13. C. Calculations • Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. GIVEN: f.p. = ? tf = ? WORK: m = 0.48mol ÷ 0.100kg 0 – tf tf = (2)(4.8m)(1.86°C/m) tf = 18°C f.p. = 0.00°C – 18°C f.p. = -18°C i = 2 m = 4.8m Kf = 1.86°C/m tf = i · m · Kf

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