Chapter 12 Topics
Physical Properties of Solutions
• Solution, solute, solvent, unsaturated, saturated, supersaturated, crystallization
•Solution Process
•Concentrations (%mass, Mole fraction, molality, molarity, ppm)
•Factors affecting Solubility (T and P)
•Colligative properties of nonelectrolyte solution:
1)Vapour pressure lowering
2)B.P. Elevation
3) F.P. Depression
4) Osmotic Pressure (  )
•Using Colligative properties to determine molar mass
•Colligative properties of electrolyte solution
Dr. Ali Bumajdad

Solution, solute, solvent, unsaturated, saturated, supersaturated, crystallization
•
of 2 or more substances
•
is(are) the substance(s) present in the smaller amount(s)
•
is the substance present in the larger amount

Solution not necessarily liquid , it could be gas or solid

•
Saturated solution: contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.
•
Unsaturated solution: contains less solute than the solvent has the capacity to dissolve at a specific temperature.
•
Supersaturated solution: contains more solute than is present in a saturated solution at a specific temperature (not stable).
Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate
•
Crystallization: dissolved solute comes out of solution and forms crystals.
solvation
Solute + solvent Solution crystallization

Solution Process
Three types of interactions in the solution process:
• solvent-solvent interaction (endothermic)
• solute-solute interaction (endothermic)
• solvent solute interaction (endothermic or exothermic)
D
H soln
>0, solution may form and maynot
D
H soln
<0, solution favourable
D
H soln
=
D
H
1
+
D
H
2
+
D
H
3

When solute-solute or solvent-solvent interaction are much more than solute-solvent interaction a solution will not form
Two substances with similar intermolecular forces are likely to be soluble in each other.
• non-polar molecules are soluble in non-polar solvents
CCl
4 in C
6
H
6
• polar molecules are soluble in polar solvents
C
2
H
5
OH in H
2
O
• ionic compounds are more soluble in polar solvents
NaCl in H
2
O or NH
3
(l)

• Miscible : two liquids completely soluble in each other in all proportions
• Solvation : ion or molecules surrounded by solvent molecules
Q) Can we form a solution of octane (C
8
H
18
) in water?
No because:
- octane not polar while water is polar
-In this case
D
H
3 is small while
D
H
1 and
D
H
2 large +ve hence,
D
H soln is large positive
Sa Ex. : Predict whether each of the following substances is more likely to dissolve in CCl
4 or in Water: C
7
H
16
, Na
2
SO
4
, HCl, I
2
.

(a) Br
2 in C
6
H
6
(b) KCl in NH
3
(c) CH
2
O in H
2
O

•
of a solution is the amount of solute present in a given quantity of solvent or solution.
mass of solute
= mass of solute + mass of solvent x 100%
(1)
= mass of solute mass of solution x 100%
A moles of A
= sum of moles of all components
(2)

(3)
= moles of solute liters of solution
moles of solute
= mass of solvent (kg)
(4)

Concentration Type Equation Equation
Mass Percentage
Mass % of component
 mass of component total mass of in solution solution

100 m %
 m a
 m a m b

.......

100
Mole Fraction
Mole fraction of component
 no .
of moles total no .
of of component moles
X a
 n a
 n n a b

.......
Molarity
Molality
Molarity
 total no .
of moles volume of of solute solution in liter
M
 n
V
L tot
Molality
 no .
of mass of moles of solvent solute in Kg n m
 m solvent kg
Unit
No unit
(%)
No unit mol/l or
M mol/kg or m
1) n= mass
M.m
2) D= mass
Vol

Sa. Ex.: A solution is made by dissolving 13.5 g glucose (C
6
H
12
O
6
) in
0.100 kg of water. What is the mass percentage of solute in this solution?
11.9%
Sa. Ex.
(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of
NaCl in 50. g of water.
(b) A commercial bleaching solution contains 3.62 mass% sodium hypochlorite,
NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?
2.91%
90.5g

Sa. Ex.
a) A solution is made by dissolving 4.35 g glucose (C
6
H
12
O
6
) in
25.0 ml of water. Calculate the molality of glucose in the solution.
0.966 m b) What is the molality of a solution made by dissolving 36.5 g of naphthalene (C
10
H
8
) in 425 g of toluene (C
7
H
8
)
0.670 m c) A solution of hydrochloric acid contains 36% HCl by mass
(i) Calculate the mole fraction of HCl in the solution .
(ii) Calculate the molality of HCl in the solution
X
HCl
= 0.22
15 m

Q) What is the molality of a 5.86 M ethanol (C
2
H
5
OH) solution whose density is 0.927 g/mL?
moles of solute moles of solute m = M = mass of solvent (kg) liters of solution
Assume 1 L of solution:
(1) mass of solute : (n

Mm=mass)
5.86
moles ethanol contains 270 g ethanol
(2) mass of solution : solution density
 solution volume
927 g of solution mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg m = moles of solute mass of solvent (kg)
=
5.86 moles C
2
H
5
OH
0.657 kg solvent
= 8.92 m

Sa. Ex. A solution contains 5.0 g of toluene (C
7
H
8
) and 225 g of benzene and has a density of 0.876 g/ml. Calculate the molarity of the solution.
0.21 M
Sa. Ex. A solution containing equal masses of glycerol (C
3
H
8
O
3
,
Mm=92gmol -1 ) and water has a density of 1.10 g/ml. Calculate:
(a) The molality of glycerol
(b) The mole fraction of glycerol
10.9 m
0.163
(c) The molarity of glycerol in the solution.
5.97 M

•
(1) Temperature
Like dissolve like a)
solubility and temperature
Usually s

T
For solid

• Fractional crystallization: is the separation of a mixture of substances into pure components on the basis of their differing solubilities.
Suppose you have 90 g KNO
3 contaminated with 10 g NaCl.
Fractional crystallization:
1.
Dissolve sample in 100 mL of water at 60 0 C
2.
Cool solution to 0 0 C
3.
All NaCl will stay in solution
(s = 34.2g/100g)
4.
78 g of PURE KNO
3 will precipitate (s = 12 g/100g).
90 g – 12 g = 78 g

(1) Temperature
solubility and temperature
Usually s

1
For gas
T

(2) Pressure a)
solubility and Pressure
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (
).
Henry’s law c = kP
(5) (there are some exceptions when the gas react with water) c is the concentration ( M ) of the dissolved gas
P is the pressure of the gas over the solution (vapor or partial pressure) k is a constant for each gas (mol/L •atm) that depends only on temperature
Vapour Pressure : pressure of gas above the liquid low P low c high P high c

Sample Ex. : Calculate the concentration of CO
2 in soft drink (partial pressure of CO
2
= 4.0 atm, and the Henery's law constant for CO is 3.1

10 -2 mol/L atm.
2 in water c = kP
(5)
=0.12 mol/l

Colligative Properties of Nonelectrolyte Solutions
• Colligative properties: properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
(e.g. adding salt and ethylene glycol reduce F.P.)
Adding nonvolatile solute to a solvent always lower the vapour pressure
P
1
= X
1
P
1
0
(6)
For nonvolatile solute (no measurable vapour pressure)
Where:
P
1
P
1
0
X
1
= vapor pressure of solution after adding solute
= vapor pressure of pure solvent
= mole fraction of the solvent

If the solution contains only one solute:
X
1
= 1 – X
2
P
1
=
2
P
1
0
P
1
0 P
1
=
D
P = X
2
P
1
0
(7)
X
2
= mole fraction of the solute
The decrease in vapour pressure (
D
P) directly proportion to solute concentration (not to the nature of solute)

When both components are volatile:1) Ideal behaviour
Ideal Solution
P
A
= X
A
P 0
A
P
B
= X
B
P 0
B
P
T
= P
A
+ P
B
P
T
= X
A
P 0
A
+ X
B
P 0
B

When both components are volatile: 2) Nonideal behaviour
P
T is greater than predicted by Raoults’s law
P
T is less than predicted by Raoults’s law
Force
A-B
<
Force
A-A
&
Force
B-B
Force
A-B
>
Force
A-A
&
Force
B-B

Used to separate two liquids from each other based on their boiling point

Vapour pressure of water at 30 °C = 31.82mmHg
P
1
= X
1
P
1
0
(6)
Raoult’s Law
P
1
0 P
1
=
D
P = X
2
P
1
0
(7)
X
1
= n
1 n
1
+n
2
(2)
X
2
=1-X
1

Sa. Ex. Glycerin (C
3
H
8
O
3
) is a nonvolatile nanelectrolyte with a density of 1.26 g/ml at 25 °C. Calculate the vapor pressure at 25 °C of a solution made by adding 50.0 ml of glycerin to 500.0 ml of water. The vapor pressure of pure water at 25 °C is 23.8 torr.
23.2 torr
Sa. Ex. The vapor pressure of pure water at 110 °C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 °C.
Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?
0.290

By the addition of nanvolatile solute D
T b
= T b
– T b
0
T b
0 is the boiling point of the pure solvent
T b is the boiling point of the solution
T b
> T b
0 D
T b
> 0
D
T b
= K b m
(8)
T b
0 T b m is the molality of the solution
K b is the molal boiling-point elevation constant ( 0 C/ m ) for a given solvent
(depend only on solvent)

By the addition solute
D
T f
= T
0 f
– T f
T f
0 is the freezing point of the pure solvent
T f is the freezing point of the solution
T f
0
> T f
D
T f
> 0
D
T f
= K f m
(9)
T f
T f
0 m is the molality of the solution
K f is the molal freezing-point depression constant ( 0 C/ m ) for a given solvent
(depend only on solvent)

The decrease in T f upon addition of solute is due to the more energy needs to be removed to form order solution than to form order solvent (solution is more disorder)
The solid that separate when a solution freeze is the pure solvent only that why the V.P. of solid is the same as that for pure liquid (see the phase diagram)

Q) What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g/mol.
(K f
0
m)
D
T f
= K f m
(9) m = moles of solute mass of solvent (kg)
=
478 g
62.01 g
3.202 kg solvent
= 2.41 m
D
T f
= K f m = 1.86 0 C/ m x 2.41 m = 4.48 0 C
D
T f
= T f
0 – T f
T f
= T f
0 – D
T f
= 0.00 0 C – 4.48 0 C = -4.48 0 C

Sa. Ex. Automotive antifreeze consists of ethylene glycol (C
2
H
6
O
2
), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass% solution of ethylene glycol in water.
D
T b
(K b
= water=0.52 and K f
K b m
(8) water = 1.86 °C/m)
D
T f
= K f m
(9)
T b
= T b
0 +
D
T b
T f
= T f
0 -
D
T b
102.8 °C -10.0 °C
Sa. Ex. Calculate the freezing point of a solution containing 0.600 kg of
CHCl
3 and 42.0 g of eucalyptol (C in the leaves of eucalyptus trees.
10
H
18
O), a fragrant substance found


• Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.
• Semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.
• Osmotic pressure (

) is the pressure required to stop osmosis.
Cucumber in concentrated brine lose water Pickle dilute more concentrated

High
P
Low
P
Osmotic pressure in unit atm

= MRT
Colligative property = depend on concentration only
M is the molarity of the solution
R is the gas constant (0.0821 L atm/mol K)
T is the temperature ( in K )
(10)

• When the concentrations of the two solution are equal and, hence, have the same osmotic pressure, the solutions called
•
: the two solutions are of equal concentration and, hence, they have the same osmotic pressure
• When concentrations of the two solution are different, the lower concentration called
and the higher concentration called
•
: the solution of lower concentration
•
: the solution of higher concentration

A cell in isotonic solution the two solution are equal in concentration
A cell in hypotonic solution
A cell in hypertonic solution the outer solution is less concentrated than the solution in the cell the outer solution is more concentrated than the solution in the cell


= MRT
(10)
1.23 M

Sa. Ex.: The average osmotic pressure of blood is 7.7 atm at 25 °C.
what concentration of glucose (C
6
H
12
O
6
) will be isotonic with blood?
0.31 M
Sa. Ex.: What is the osmotic pressure at 20 °C of a 0.0020 M sucrose (C
12
H
22
O
11
) solution?
0.048 atm

• Using Colligative Properties to determine molar mass
B.P. Elevation
D
T b
= K b m
(8)
F.P. Depression
D
T f
= K f m
(9) m =
D
T b
K b m =
D
T f
K f m = n mass of solvent in Kg m = n mass of solvent in Kg m = mass in g
M.m. (mass of solvent in Kg)
M.m
= mass in g
(11) m (mass of solvent in Kg) m = mass in g
M.m. (mass of solvent in Kg)
M.m
= mass in g
(11) m (mass of solvent in Kg)

Osmotic Pressure

= MRT
(10)
M =

RT
M = n
V
L
M = mass in g
M.m. (V
L
)
M.m
= mass in g
M (V
L
)
M.m
= ( mass in g) R T
(V
L
)

(12)
(13)

K f of benzene = 5.12 °C/ m m =
D
T f
K f
M.m
= mass in g
(11) m (mass of solvent in Kg)
127 g/mol
C
10
H
8

1 atm = 760 mmHg
6.51

10 4 g/mol

Sa. Ex.: Coamphor (C10H16O) melts 1t 179.8 °C, and it has a particularly large freezing-point-depression constant Kf = 40.0
°C/ m . When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 °C. what is the molar mass of the solute?

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
P
1
= X
1
P
1
0 (7)
D
T b
= K b m
D
T f
= K f m


= MRT
(8)
(9)
(10)

0.1 m NaCl solution 0.1 m Na + ions & 0.1 m Cl ions
• Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
0.1 m NaCl solution 0.2 m ions in solution van’t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln nonelectrolytes
NaCl
CaCl
2 i should be
1
2
3

D
T b
= i K b m
(8b)
D
T f
= i K f m
(9b)


= iMRT (10b)


= i MRT

Sa. Ex.: List the following solutions in the order of their expected freezing points: 0.050 m CaCl
2
; 0.15 m NaCl; 0.10 m HCl;
0.050 m HC
2
H
3
O
2
; 0.10 m C
12
H
22
O
11
.
Sa. Ex: Which of the following solutes will produce the largest increase in boiling point upon addition to 1 Kg of water:
1 mol Co(NO
3
)
2
, 2 mol of KCl, 3 mol of ethylene glycol (C
2
H
6
O
2
)?