1. Colligative Properties (solutions)
A. Definition
Colligative Property
property that depends on the concentration of solute particles, not their identity

2. 4 Colligative Properties
Vapor-pressure
Freezing point
Boiling point
Osmotic pressure

3. Vapor pressure
vapor pressure is the measure of the tendency of molecules to escape from a liquid
volatile- high vapor pressure
non-volatile- low vapor pressure
vapor pressure lowering depends on the concentration of a nonelectrolyte solute and is independent of solute identity, it is a colligative property
because vapor pressure is lowered, this lowers the freezing point and raises the boiling point.

4. Freezing Point Depression
(tf) - is the difference between the freezing points of the pure solvent, and it is directly proportional to the molal concentration of the solution.
f.p. of a solution is lower than f.p. of the pure solvent

5. the freezing point of a 1-molal solution of any nonelectrolyte solute in water is found by experiment to be 1.86°C lower than the freezing point of water.
Molal freezing-point constant, K, is the freezing point depression of the solvent in a 1- molal solution of a nonvolatile, nonelectroyte solute
Each substance has a different molal freezing point constant (p. 438)

6. Applications salting icy roads making ice cream antifreeze
cars (-64°C to 136°C)

7. ∆tf = Kf mi
∆tf = freezing point depression (°C)
Kf = °C/ m m = mol solute/kg of solvent
i = # of particles

8. Boiling Point Elevation
(tb)
b.p. of a solution is higher than b.p. of the pure solvent
When the vapor pressure is equal to the atmospheric pressure boiling will occur.
molal boiling point constant (kb ) is the boiling-point elevation of the solvent in a 1-molal solution of a nonvolitle, nonelectrolyte solute. (0.51 °C/m)

9. Boiling-point elevation,∆tb , is the difference between the boiling points of the pure solvent and a nonelectrolyte solution of that solvent, and is directly proportional to the molal connection of the solution.

10. ∆tb = Kb m i
∆tb = boiling-point elevation (°C)
Kb = °C/ m
m = mol solute/kg of solvent
i = # of particles

11. Osmotic pressure:
Semipermeable membranes allow the movement of some particles while blocking the movement of others.
Osmosis: the movement of solvent through a semipermeable membrane from the side of lower solute concentration to the side of higher solute concentration.
Osmotic pressure is the external pressure that must be applied to stop osmosis.

12. Electrolytes: remember dissociation of ionic compounds
NaCl lowers the freezing point twice as much as sucrose C12H22O11
NaCl Na+ + Cl-
CaCl2 lowers the freezing point three times as much as C12H22O11 due to the dissociation of the ionic compounds
CaCl2 → Ca Cl-

13. C. Calculations t: change in temperature (°C)
t = k · m · i
t: change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
(different depending on freezing or boiling)
m: molality (m)
i: # of particles

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

15. When we dissolve most ionic substances in water they break up into their individual ions.
NaCl(s)  Cl- (aq) + Na+ (aq)
Most molecules don’t break up into ions. For example sugar.
C12 H24O12 (s) + H2O (aq)  C12 H24O12 (aq)
But some molecules do break up into ions. Acids and bases are examples.
HCl(l) + H2O (aq)  Cl- (aq) + H3O + (aq)

16. Dissociation Equations ( ionic )
NaCl(s) 
Na+(aq) + Cl-(aq)
i = 2
AgNO3(s) 
Ag+(aq) + NO3-(aq)
i = 3
MgCl2(s) 
Mg2+(aq) + 2 Cl-(aq)
i = 3
Na2SO4(s) 
2 Na+(aq) + SO42-(aq)
AlCl3(s) 
Al3+(aq) + 3 Cl-(aq)
i = 4

17. C. Calculations GIVEN: WORK: b.p. = ? m = 0.73mol ÷ 0.225kg tb = ?
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
i = 1
tb = kb · m · i

18. C. Calculations GIVEN: WORK: f.p. = ? m = 0.48mol ÷ 0.100kg tf = ?
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
i = 2
tf = kf · m · i

19. What is the boiling point?
What is the boiling point of a solution made by dissolving 1.20 moles of NaCl in 750 g of water?
What is the boiling point?
Δtb = kb .m.i
Molality=1.20moles/.750 kg
molality= 1.6 m
Δtb = (0.51°C/m)(1.6m)(2)
Δtb = 1.63°C
Boiling pt: = 102 °C
Find molality!

20. Molar mass of sucrose is 342.34 g/mol Find moles: = 0.04995 mol
Ex: What is the freezing point depression of water in a solution of 17.1 g of sucrose, C12 H22O11 , and 200. g of water? What is the actual freezing point of the solution?
Δtf = kf .m.i
Molar mass of sucrose is g/mol
Find moles:
= mol
Molality= moles/.200 kg
molality= m
Δtb = (-1.86°C/m)(0.2498m)(1)
Δtb = °C

21. Ex: A water solution containing an unknown quantity of a nonelectrolyte solute is found to have a freezing point of
-0.23°C. What is the molal concentration of the solution?
Δtf = kf .m.i
-0.23 °C = (-1.86 °C/m)(x)(1)
-0.23 °C = x
-1.86 °C
0.12 m = x

22. Find molality first: 20.0 g / 62.0 g/mol
Ex: What is the boiling point elevation of a solution made from 20.0 g of a nonelectrolyte solute and g of water? The molar mass of the solute is 62.0g/mol.
Find molality first: 20.0 g / 62.0 g/mol
Molality = mol/0.400 kg=0.806 m
Δtb = kb .m.i
= (0.51 °C/m)(0.806 m)(1)
= °C

23. Ex: What is the expected change in the freezing point of water in a solution of 62.5 g of barium nitrate, Ba(NO3 )2 , in 1.00 kg of water?
Molality= 62.5 g/ g/mol = m
Δtf = kf .m.i
= (-1.86 °C/m)( m)(3)
= °C

24. Assignment: Do practice problems 1-4 on page 440
Practice problems have answers off to the right.