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
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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 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. III. Colligative Properties
Ch 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. 2. Freezing Point Depression
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. 3. Boiling Point Elevation
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. 0.51 degrees Kelvin for every Mole of solute added to 1KG of Water
C. Calculations
tf(b) = kf(b) · m · n
t: change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
m: molality (m)
n: # of particles
0.51 degrees Kelvin for every Mole of solute added to 1KG of Water

16. C. Calculations NaCl Na+ + Cl- n=2 MgCl2 Mg+2 + 2Cl- n=3
# of Particles
Nonelectrolytes (covalent)
remain intact when dissolved
1 particle (C6H12O C6H12O n=1)
Electrolytes (ionic)
dissociate into ions when dissolved
2 or more particles
NaCl Na Cl n=2
MgCl Mg Cl- 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 p = MRT
The pressure required to prevent osmosis is known as osmotic pressure (p) of the solution.
p = MRT
M = Molarity
R = 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 kg of water. The boiling point of the solution is elevated C 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 = °C
kb = 0.51°C·kg/mol
WORK:
.433°C=(0.51°C·kg/mol)(Xmol) · (1 ) kg
X = moles
MM= mass/moles = 10.0g/.0849 mol
MM= 118 g/mol
m = ? Moles/kg
Kg solv = kg
n = 1
tb = kb · x moles/kg · n