1. 13.5 Colligative properties
Dissolving solute in pure liquid will change all physical properties of liquid, Density, Vapor Pressure, Boiling Point, Freezing Point, Osmotic Pressure
Colligative Properties are properties of a liquid that change when a solute is added.
The magnitude of the change depends on the number of solute particles in the solution, not on the identity of the solute particles.

2. Lowering the vapor pressure
A liquid in a closed container will establish equilibrium with its vapor. When that equilibrium is reached, the pressure exerted by the vapor is called the vapor pressure
The presence of a non-volatile solute means that fewer solvent particles are at the solution’s surface, so less solvent evaporates

3. Raoult’s Law Describes vapor pressure lowering mathematically .
The lowering of the vapour pressure when a non-volatile solute is dissolved in a volatile solvent (A) can be described by Raoult’s Law:
PA =XAPA
PA = vapour pressure of solvent A above the solution
XA = mole fraction of the solvent A in the solution
PA = vapour pressure of pure solvent A
only the solvent (A) contributes to
the vapour pressure of the solution

4. ( ( ( P =X P P =(0.987)(23.76 mmHg) = 23.5 mmHg
Example: What is the vapor pressure of water above a sucrose (MW=342.3 g/mol) solution prepared by dissolving g of sucrose in g of water at 25 ºC? The vapor pressure of pure water at 25 ºC is mmHg.
P =X P
soln
H2O (
(158 g C12H22O11)
1 mol C12H22O11
342.3 g C12H22O11
Moles C12H22O11 =
= mol (
( g H2O)
1 mol H2O
18 g H2O
Moles H2O =
= 35.6 mol (
mol H2O
mol H2O + mol C12H22O11
H2O = X
35.6 =
= 0.987
P =(0.987)(23.76 mmHg) = 23.5 mmHg
soln

5. Example: Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25C of solution made by adding mL of glycerin to mL of water. The vapor pressure of pure water at 25C is 23.5 torrr and its density is 1.0 g/mL. Calculate the vapor pressure lowering
P =X P
H2O (
(50 mL C3H8O3)
1.26 g C3H8O3
1 mL C3H8O3 (
1 mol C3H8O3
92.1 g C3H8O3
Moles C3H8O3 =
= mol (
(500 mL H2O)
1.0 g H2O
1 mL H2O (
1 mol H2O
18 g H2O
Moles H2O =
= 27.8 mol (
mol H2O
mol H2O + mol C3H8O3 (
27.8
H2O = X =
= 0.976
P =(0.976)(23.8 torr) = 23.2 torr
H2O

6. Example: The vapor pressure of pure water at 110C is 1070 torr
Example: The vapor pressure of pure water at 110C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of atm at 110C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?
P =X P
H2O
H2O
P = 1070 torr
H2O
P = 1 atm = 760 torr
H2O P
H2O = X P =
760 torr
1070 torr
= 0.710 X
H2O
C2H6O2 +
= 1 X
C2H6O2
= 1  0.71 = 0.290

7. Mixtures of Volatile Liquids Both liquids evaporate & contribute to the vapor pressure

8. Raoult’s Law: Mixing Two Volatile Liquids
Since both liquids are volatile and contribute to the vapour, the total vapor pressure can be represented using Dalton’s Law:
PT = PA + PB
The vapor pressure from each component follows Raoult’s Law:
PT = XAPA + XBPB 
Also, XA + XB = (since there are 2 components)

9. Benzene and Toluene A two solvent (volatile) system
The vapor pressure from each component follows Raoult's Law.
Benzene - Toluene mixture:
Recall that with only two components, XBz + XTol = 1
Benzene: when XBz = 1, PBz = PBz = 384 torr & when XBz = 0 , PBz = 0
Toluene: when XTol = 1, PTol = PTol = 133 torr & when XTol = 0, PBz = 0  

10. PT = [(0.33)(75 torr)] + [(0.67)((22 torr)]
Example: A mixture of benzene (C6H6) and toluene (C7H8) containing 1.0 mol of benzene and 2.0 mol of toluene. What is the total vapor pressure of the solution? [vapor pressures of pure benzene and toluene are 75 torr and 22 torr, respectively]
PT = XAPA + XBPB 
C6H6 =
= 0.33 1 X
1+2
C7H8 =
= 0.67 2 X
1+2
PT = [(0.33)(75 torr)] + [(0.67)((22 torr)]
= torr torr
= 39.5 torr

11. Boiling-point elevation and Freezing-point depression
In a solution of a nonvolatile solute, boiling and freezing points differ from those of the pure solvent
Boiling point is elevated when solute inhibits solvent from escaping.
The boiling point of the solution is higher than that of the pure liquid
Freezing point is depressed when solute inhibits solvent from crystallizing.
The freezing point of the solution is lower than that of the pure liquid

12. Notice the changes in the boiling & freezing points.
The diagram below shows how a phase diagram is affected by dissolving a solute in a solvent.
The black curve represents the pure liquid and the blue curve represents the solution.
Notice the changes in the boiling & freezing points.
Phase diagrams for a pure solvent and for a solution of nonvolatile solute

13. Boiling-point elevation
The increase of boiling point, Tb is directly proportional to the concentration of the solution expressed by its molality, m.
Tb = (Tb –Tb ) = kbm 
Where, Tb = BP. Elevation
Tb = BP of solvent in solution
Tb° = BP of pure solvent
m = molality , kb = BP Constant
The boiling-point elevation is proportional to the concentration of solute particles, regardless of whether the particles are molecules or ions
A 1 m aqueous solution of NaCl is 1 m Na+ and 1 m Cl-, making 2 m in total solute particles
The boiling-point of elevation of a 1 m aqueous solution of NaCl is (2m)(0.51 C/m) = 1C.

14. Freezing-point depression
The decrease of freezing point, Tf is directly proportional to the concentration of the solution expressed by its molality, m.
Tf = (Tf –Tf) = kfm 
Where, Tf = FP depression
Tf = FP of solvent in solution
Tf°= FP of pure solvent
m = molality
kf = FP depression constant

15. The van 't Hoff factor, i : The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved, and the concentration of a substance as calculated from its mass.
For most non-electrolytes dissolved in water, the van' t Hoff factor is essentially 1.
For most ionic compounds dissolved in water, the van't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
Tf (calculated for nonelectrolyte)
i =
Tf (measured)
For NaCl, van’t Hoff factor is 2, because NaCl consists of one Na+ and on Cl- per formula unit

16. Example: Automotive antifreeze consists of ethylene glycol, (CH2(OH)CH2(OH), a nonvolatile noneletrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene glycol in water.[kb=0.51 (C/m) and kf=1.86 (C/m).
Let us consider we have 1000 g of solution:
Mass of ethylene glycol = 250 g
Mass of water = 750 g (
moles C2H6O2
Kg H2O (
250 g C2H6O2
750 g H2O = (
1 mol C2H6O2
62.1 g C2H6O2 (
1000 g H2O
1 kg H2O
Molality=
= 5.37 m
Tb = (Tb –Tb ) = kbm 
Tf = (Tf –Tf ) = kfm 
= (0.51 C/m )  (5.37 m)
= (1.86 C/m )  (5.37 m)
= 2.7 C
= 10.0 C
Tb=Tb +Tb = 2.7 C +100 C 
= C
Tf=Tf – Tf = 0 C – 10 C 
= – 10.0 C

17. Example: Calculate the freezing point of a solution containing 0
Example: Calculate the freezing point of a solution containing kg of CHCl3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees [kf=4.68 (C/m) and Tf=  63.5 C for chloroform]. (
moles C10H18O
Kg CHCl3 (
42 g C10H18O
0.60 kg CHCl3 = (
1 mol C10H18O
154 g C10H18O
Molality=
= 0.45 m
Tf = (Tf –Tf ) = kfm 
= (4.68 C/m )  (0.45 m)
= 2.1 C
Tf =Tf – Tf = – 63.5 C – 2.1 C 
= – 65.6 C

18. Example: List the following aqueous solutions in order of their expected freezing point :
0.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, m CH3COOH, 0.10 m C12H22O11
CaCl2, NaCl and HCl are stronger electrolytes
CH3COOH is week electrolyte
C12H22O11 is nonelectrolyte
0.050 m CaCl2  m in Ca2+ and 0.10 m in Cl-  m in particles
0.15 m NaCl  0.15 m in Na+ and 0.15 m in Cl-  m in particles
0.10 m HCl  0.10 m in H+ and 0.10 m in Cl  0.20 m in particles
0.050 m CH3COOH  between m and 0.10 m in particles
0.050 m C12H22O11  m in particles
0.15 m NaCl (lowest freezing-point ), 0.10 m HCl , m CaCl2, m C12H22O11, m CH3COOH (highest freezing-point )

19. Osmosis
Osmosis is the spontaneous movement of water across a semi-permeable membrane from an area of low solute concentration to an area of high solute concentration
Osmotic Pressure - The pressure that must be applied to stop osmosis

20. The osmotic pressure obeys a law similar in form to the ideal gas law
 V= nRT
 =(n/V)RT = MRT
, osmotic pressure of soln
V, volume soln
n, number of moles of solute
R, ideal-gas constant
M, molarity of soln
Two solutions having identical osmotic pressure are isotonic
Solution of lower osmotic pressure is hypotonic with respect to more concentrated soln
Solution of more concentrated solution is hypertonic with respect to the dilute soln

21. Example: The average osmotic pressure of blood is 7. 7 atm at 25 C
Example: The average osmotic pressure of blood is 7.7 atm at 25 C. What molarity of glucose (C6H12O6) will be isotonic with blood?
 = M RT
T = = 298 K
R = L.atm/mol.K
M =  RT
 = 7.7 atm
M = ?
M =
( L.atm/mol.K)(298 K)
7.7 atm
= 0.31 atm

22.  = (0.0020 (mol/L)) (0.0821 L.atm/mol.K)(293 K)
Example: What is the osmotic pressure at 20 C of a M sucrose (C12H22O11) solution?
T = = 293 K
R = L.atm/mol.K
 = M RT
M = M = (mol/L)
 = ?
 = ( (mol/L)) ( L.atm/mol.K)(293 K)
= atm

23. The colligative properties of solutions provides a useful means experimentally determining molar mass.
Example: A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.25 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was C higher than that of the pure solvent. Calculate the molar mass of the solute.
Tb = kbm
Tb Kb =
0.357 C
5.02 C/m
Molality =
= m (
mol solute
Kg CCl4
(0.040 kg CCl4)
Number of mole of solute in the solution =
= mol solute (
0.25 g
mol
Molar mass =
= 88 g/mol

24. ( ( Tf = kfm = 0.0775 m = Molality = Tf Kf 3.1 C 40.0 C/m
Example: Camphor (C10H16O) melts at C, and it has a particularly large freezing-point-depression constant, kf =40.0 C/m. When g of an organic substance of unknown molar mass is dissolved In g of liquid camphor, the freezing point of the mixture is found to be C. What is the molar mass of the solute?
Tf = – = 3.1 C
Tf = kfm
= m =
Molality =
Tf Kf
3.1 C
40.0 C/m
Number of mole of solute in the solution = (
mol solute
Kg C10H16O
( kg C10H16O)
= mol solute
Molar mass = (
0.186 g
mol
= 110 g/mol

25. Example: The osmotic pressure of an aqueous solution of a certain protein was measured to determine the protein’s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25C was found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass.
T = = 298 C
R = L.atm/mol.K
 = 1.54 torr =1.54/ = atm
 = M RT
atm
( L.atm/mol.K )(298 K)
Molarity =
= 8.28  10-5 mol /L
Mole of solute in the solution =
(5.00  10-3 L)
(8.28  10-5 mol /L)
= 4.14  10-7 mol
Molar mass = ( g
4.14  10-7 mol
= 8.45  103 g/mol

26. Example: A sample of 2.05 g of polystyrene of uniform polymer chain length was dissolved in enough toluene to form L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25C. Calculate the molar mass of the polystyrene.
T = = 298 C
R = kg.m2/S2.mol.K
 = 1.21 kPa=1210/ = atm
 = M RT
atm
( L.atm/mol.K )(298 K)
Molarity =
= 4.88  10-4 mol /L
Mole of solute in the solution =
(0.10 L)
(4.88  10-4 mol /L)
= 4.88  10-5 mol
Molar mass = (
2.05 g
4.88  10-5 mol
= 4.20  104 g/mol