1. II III I IV. Colligative Properties of Solutions Ch. 16 – Mixtures & Solutions

2. A. Definition Colligative Property Colligative Property property that depends on the number of solute particles, not their identity in an ideal solution

3. B. Types Freezing Point Depression Freezing Point Depression ( T f ) f.p. of a solution is lower than f.p. of the pure solvent Boiling Point Elevation Boiling Point Elevation ( T b ) b.p. of a solution is higher than b.p. of the pure solvent

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

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

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

7. C. Calculations t :change in temperature (° C ) k :constant based on the solvent (° C·kg/mol ) m :molality ( m ) n :# of particles t = k · m · n

8. C. Calculations T : change in temperature (° C ) i : Vant 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

9. C. Calculations # of Particles # of Particles Nonelectrolytes (covalent) remain intact when dissolved 1 particle Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles

10. C. Calculations At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? WORK: GIVEN:

11. C. Calculations At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? m = 3.2m n = 1 t b = k b · m · n WORK: m = 0.73mol ÷ 0.225kg GIVEN: b.p. = ? t b = ? k b = 3.04°C·kg/mol t b = (3.04°C·kg/mol)(3.2m)(1) t b = 9.7°C b.p. = 181.8°C + 9.7°C b.p. = 192°C

12. C. Calculations Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. WORK: GIVEN:

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

14. C. Calculations T Change in temperature Not actual freezing point or boiling point Change from FP or BP of pure solvent Freezing Point (FP) i T F is always subtracted from FP of pure solvent Boiling Point (BP) i T B is always added to BP of pure solvent

15. C. Calculations i – VHF i – VHF Nonelectrolytes (covalent) remain intact when dissolved 1 particle Electrolytes (ionic) dissociate into ions when dissolved number of ions per formula unit 2 or more particles

16. C. Calculations i – VHF i – VHF Examples CaCl 2 Ethanol C 2 H 5 OH Al 2 (SO 4 ) 3 Methane CH 4 i = 3 1 5 1

17. C. Calculations K – molal constant K – molal constant K F K F – molal freezing point constant Changes for every solvent 1.86 °C·kg/mol (or °C/ m ) for water K B K B – molal boiling point constant Changes for every solvent 0.512 °C·kg/mol (or °C/ m ) for water

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

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

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

21. D. Osmotic Pressure Osmosis: The flow of solvent into a solution through a semipermeable membrane Semipermeable Membrane: membrane that allows solvent to pass through but not solute

22. D. Osmotic Pressure Net transfer of solvent molecules into the solution until the hydrostatic pressure equalizes the solvent flow in both directions

23. Because the liquid level for the solution is higher, there is greater hydrostatic pressure on the solution than on the pure solvent Osmotic Pressure: The excess hydrostatic pressure on the solution compared to the pure solvent D. Osmotic Pressure

24. Osmotic Pressure: Minimum Pressure required to stop flow of solvent into the solution D. Osmotic Pressure

25. Osmosis at Equilibrium

26. = i M R T where: osmotic pressure (atm) π = osmotic pressure (atm) VHF i = VHF M = Molarity (moles/L) R = Gas Law Constant T = Temperature (Kelvin) E. Osmotic Pressure Calculations 0.08206 L atm/mol K

27. E. Osmotic Pressure Calculations Calculate the osmotic pressure (in torr) at 25 o C of aqueous solution containing 1.0g/L of a protein with a molar mass of 9.0 x 10 4 g/mol. i = 1 M = 1.11 x 10 -5 M R = 0.08206 L atm/mol K T = 25 o C = 298 K WORK: M = 1.0 g prot. GIVEN: = ? 1.11 x 10 -5 M = (1)(1.11x10 -5 )(.08206)(298) = 2.714 x 10 -4 atm = 0.21 torr 1 mol prot. 1 L soln 9.0 x 10 4 g =

28. Colligative Properties useful for: characterizing the nature of a solute after it is dissolved in a solvent determining molar masses of substances D. Osmotic Pressure

29. If the external pressure is larger than the osmotic pressure, reverse osmosis occurs One application is desalination of seawater F. Reverse Osmosis

30. Net flow of solvent from the solution to the solvent

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