Colligative Properties in Chemistry: Vapor Pressure, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pre

1. Note – Only 22 days until all Missing and Late Work is due by 5/27/16 unless previous arrangements have been made with me. Thank You

2. Bellringer • Suppose someone offered to give you a 3 gallon bucket filled with gold if you carry the bucket of gold up a flight of stairs at Rocky Mountain high. But if you fail to carry the bucket up the flight of stairs, you must stay in Chemistry forever. Would you accept the offer? How much gold would you be carrying in pounds? There are 72.99 kilograms in 1 gallon at a price (today) $39.79/gram down 16 cents from yesterday. Calculated your new net worth and how much money you potentially lost with the drop in gold prices from yesterday.

3. Ch. 13 - Solutions III. Colligative Properties

4. A. Definition • Colligative Property • Property that depends on the concentration (# of moles) of solute particles, not their chemical nature (identity).

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

6. B. Types • 1. Vapor Pressure Reduction VAPOR PRESSURE is the pressure of gas particles of the liquid over the liquid. The more volatile the more vapor pressure. (2nd one) View Flash animation.

7. B. Types The addition of a non-volatile solute lowers the rate of evaporation, it means lowers the vapor pressure.

8. B. Types • The magnitude of vapor pressure reduction is proportional to the solute concentration. • This is called Raoult’s Law.

9. B. Types • Raoult’s Law: • Psoln = XsolvPsolv + XsoluPsolu but if the solute is nonvolatile (p=0) we don’t need this

10. B. Types 2. Freezing Point Depression

11. 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

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

13. B. Types 3. Boiling Point Elevation Solute particles weaken IMF in the solvent.

14. B. Types • 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

15. 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 degrees Kelvin for every Mole of solute added to 1KG of Water

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

17. C. Calculations • At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil?

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

19. C. Calculations • At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? GIVEN: b.p. = ? tb = ? kb = 3.60°C·kg/mol WORK: m = 0.73mol ÷ 0.225kg tb = (3.60°C·kg/mol)(3.2m)(1) tb = 12°C b.p. = 181.8°C + 12°C b.p. = 194°C m = 3.2m n = 1 tb = kb · m · n

20. C. Calculations • Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.

21. 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·kg/mol WORK: m = 0.48mol ÷ 0.100kg tf = (1.86°C·kg/mol)(4.8m)(2) tf = 18°C f.p. = 0.00°C - 18°C f.p. = -18°C m = 4.8m n = 2 tf = kf · m · n

22. 4. OSMOTIC PRESSURE Osmosis: The movement of a solvent through a semipermeable (some molecules can pass) membrane from a dilute solution (less concentrated) to a more concentrated one (flows from low concentration to high concentration).

23. 4. OSMOTIC PRESSURE The pressure required to prevent osmosis is known as osmotic pressure (p) of the solution. p = MRT M = Molarity R = 0.082 l.atm/mol.K T = Temp. in K.

24. D. Molar Mass determination • Any colligative property can be used to determine the molar mass of an unknown • Molar Mass = mass of unknown/# moles

25. D. Molar Mass determination • A 10.0 g sample of an unknown compound that does not dissociate is dissolved in 0.100 kg 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?

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

27. D. Molar Mass determination Use: tb = kb · m · n to find # of moles GIVEN: tb = kb = WORK: tb = kb · x moles/kgsolv · n m = ? Moles/kg Kg solv = n =

28. D. Molar Mass determination Use: tb = kb · m · n to find # of moles GIVEN: tb = 0.433 °C kb = 0.51°C·kg/mol WORK: .433°C=(0.51°C·kg/mol)(Xmol) · (1 ) 0.100 kg X = 0.0849 moles MM= mass/moles = 10.0g/.0849 mol MM= 118 g/mol m = ? Moles/kg Kg solv = 0.100 kg n = 1 tb = kb · x moles/kg · n