CHAPTER 4
First year
 Solutions
C
By
Dr. Hisham Ezzat
2011- 2012
1
Completely miscible liquids
1.
–
–
2.
3.
Ideal solution
Non - ideal solution
Completely immiscible liquids H2O and
aniline, H2O and chlorobenzene
Partially immiscible liquids, H2O and
phenol, H2O and ether
Ideal Solution
1.
2.
3.
4.
The force of attraction between all molecules
are identical i.e. the attraction force is not
affected by addition of other components A A = B-B = A - B.
No heat is evolved or absorbed during mixing
i.e.  H soln. = Zero
The volume of solution is the sum of volumes
of the two liquids.
The solution obeys Raoult's law.
Figure (1): Vapor pressure of ideal solutions
Example 6:

Heptane (C7H16) and octane (C8H18) form
ideal solutions What is the vapor pressure
at 40°C of a solution that contains 3.0 mol
of heptane and 5 mol of octane? At 40°C,
the vapor pressure of heptane is 0.121 atm
and the vapor pressure of octane is 0.041
atm.
Solution:
The total number of moles is 8.0. therefore
X heptane = 3.0/8.0 = 0.375
X octane = 5.0/8.0 = 0.625
Total = X heptane . Po heptane + X octane.
Po octane
= 0.375 x 0.12 +0.625 x 0.04
= 0.045 atm + 0.026 atm.
= 0.071 atm.
Example 7:

Assuming ideality, calculate the vapor
pressure of 1.0 m solution of a non volatile, on dissociating solute in water at
50°C. The vapor pressure of water 50°C is
0.122 atm.
Solution :


From example 2 the mole fraction of water
in 1.0m solution is 0.982.
PH2O = XH2O PH2O = 0.982 x 0.122 = 0.120
atm.
Problem:

At 140°C, the V.P of C6H5CI is 939.4 torr
and that of C6H5Br is 495.8 torr. Assuming
that these two liquids from an ideal
solution. Find the composition of a mixture
of two liquids which boils at 140°C under 1
atm pressure?
Non-
ideal solutions
Negative deviation
Positive deviation
1- The force of attraction
The force of attraction
increase by mixing
decrease by mixing A-A , B-B
A - A, B-B < A-B
> A-B
2- The vapor pressure will be lower
The vapor pressure will be
than that given by Roault's law
higher than that given by
Raoult's law.
3- H solution :- Ve (exothermic)
 H solution: + Ve
(endothermic)
4- Temperature change when
Temperature change when
solution is formed: increase
solution is formed: decrease.
5- Example: Acetone-water
Ethanol-hexane
Fig.2: Vapour pressure of non-ideal Fig.3: Vapour pressure of non-ideal
solution (-ve deviation)
solution (+ve deviation)
Fractional Distillation of Binary
Miscible liquids
•
•
The separation of mixture of volatile liquids into their
components is called fractional distillation,
the distillate containing the more volatile component
and the residue the less volatile one
a) Ideal solutions
•
If a mixture of 2 liquids (A and B) form a
completely miscible ideal solution and PA > PB
result in B.P. of A < B.P of B thus on boiling:–
–
•
1) The Liquid A boils at lower B.P than that of liquid B.
2) The liquid A which is more volatile will be passed
from the fractionating column and the liquid B which is
less volatile returned again to the distallating flask.
A solution of intermediate B.p. between 2 pure
liquid -called azeotropic solution
b) Non - ideal solutions
(solutions that exhibit deviations from Raoults law)
Non - ideal solutions with minimum boiling point:
If a solution having any other compositions is distilled, the azeotropic mixture will
distill first and the excess of (A) or (B) will remains in the flask e.g 95 % ethanol and 5
% H2O.
2) Non - ideal solutions with maximum boiling point:
• If a solution having any other composition is distilled,
the execs of acetone or CHCI3 will distill first leaving the
azeotropic mixture in the flask.
Example 8:

A solution is prepared by mixing 5.81 g
acetone C3H6O, (M. wt = 58.1 g/mole) 11.9 g
chloroform (CHCI3 M.wt 119.4 g/mole). At
35°C this solution has a total vapor pressure
of 260 torr. Is this an ideal solution?
Comment? The vapor pressure of pure
acetone and pure CHCI3 at 35°C are 345 and
293 torr, respectively.

Colligative Properties of Solutions
20
Colligative Properties of Solutions
There are four common types of colligative
properties:

1.
2.
3.
4.

Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic pressure
Vapor pressure lowering is the key to all
four of the colligative properties.
21
Lowering of Vapor Pressure and
Raoult’s Law

Addition of a nonvolatile solute to a solution
lowers the vapor pressure of the solution.



The effect is simply due to fewer solvent
molecules at the solution’s surface.
The solute molecules occupy some of the spaces
that would normally be occupied by solvent.
Raoult’s Law models this effect in ideal
solutions.
22
Lowering of Vapor Pressure and
Raoult’s Law

Derivation of Raoult’s Law.
0
Psolvent  X solvent Psolvent
where Psolvent  vapor pressure of solvent in solution
0
Psolvent
 vapor pressure of pure solvent
X solvent  mole fraction of solvent in solution
23
Lowering of Vapor Pressure and
Raoult’s Law

Lowering of vapor pressure, Psolvent, is defined as:
Psolvent  P
0
solvent
 Psolvent
0
0
 Psolvent
- ( X solvent)( Psolvent
)
 (1  X solvent)P
0
solvent
24
Lowering of Vapor Pressure and
Raoult’s Law


Remember that the sum of the mole fractions
must equal 1.
Thus Xsolvent + Xsolute = 1, which we can
substitute into our expression.
X solute  1 - X solvent
0
Psolvent  X solute Psolvent
which is Raoult' s Law
25
Lowering of Vapor Pressure and
Raoult’s Law

This graph shows how the solution’s vapor pressure
is changed by the mole fraction of the solute, which
is Raoult’s law.
26
Examples
The vapor pressure of water is 17.5 torr at 20°C.
Imagine holding the temperature constant while
adding glucose, C6H12O6, to the water so that the
resulting solution has XH2O = 0.80 and XGlu =
0.20. What is , the vapor pressure of water over the
solution
0
PA  X A PA
PA  X P  0.80 X 17.5torr
0
A A
= 14 torr
27
Glycerin, C3H8O3, is a nonvolatile nonelectrolyte
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
28
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? Answer: 0.290
P°H2O =1070 torr
PH2O = 1 Atm = 760 torr
PH2O
torr
XH2O = --------- = 760
--------P°H2O 1070 torr =
XH2O + XEG = 1
0.7103 + XEG = 1
0.71028
1- 0.7103 = XEG
XEG =
0.28972
= 0.290
More Examples
Sucrose is a nonvolatile, nonionizing solute in
water. Determine the vapor pressure lowering, at
27°C, of a solution of 75.0 grams of sucrose,
C12H22O11, dissolved in 180. g of water. The
vapor pressure of pure water at 27°C is 26.7 torr.
Assume the solution is ideal.
1 mol Suc
n Suc  75.0 gSuc
 0.219mol
342.3g Suc
1 mol Water
nW ater  180 gWater
 9.99mol
18 g Watyer
nwater
9.991
X W ater 

 0.978541
nW ater  nUc 9.991  0.2191
PW ater  PW0ater X W ater  26.7 torr
X 0.97854  26.13
Vapor Pressure Lowered = 26.7-26.1= 0.6
30
solution is made by mixing 52.1 g of propyl
chloride, C3H8Cl, and 38.4 g of propyl bromide,
C3H8Br. What is the vapor pressure of propyl
chloride in the solution at 25°C? The vapor pressure
of pure propyl chloride is 347 torr at 25°C and that
of pure propyl bromide is 133 torr at 25°C. Assume
that the solution is an ideal solution.
1 mol CP
nCP  52.1 g CP
 0.6633
78.54 g CP
1 mol CB
nCB  38.4 g CB
 0.312
122.99 g CB
nPCr
0.6633
X PC 

 0.67996
nPC  nPB 0.6633  0.3122
0
PPC  PPC
X PC  347 X 0.679964  235.95  236Torr
31
. At 25°C a solution consists of 0.450 mole of
pentane, C5H12, and 0.250 mole of cyclopentane,
C5H10. What is the mole fraction of cyclopentane in
the vapor that is in equilibrium with this solution?
The vapor pressure of the pure liquids at 25°C are
451 torr for pentane and 321 torr for cyclopentane.
Assume that the solution is an ideal solution.
0
PPen  PPen
X Pen  .450 X 451  202.95
0
PCPen  PCPen
X CPen  0.250 X 321  80.25
PPenV
PCPenV
PV
n
; nPen 
; nCPen 
RT
RT
RT
PCPenV
nCPen
PCPen
RT
X CPen 


nCPen  n Pen PCPenV PPenV PCPen  PPen

RT
RT
80.25
PCPen 
 0.283
80.25  202.95
32
Boiling Point Elevation

Addition of a nonvolatile solute to a solution
raises the boiling point of the solution above
that of the pure solvent.



This effect is because the solution’s vapor
pressure is lowered as described by Raoult’s law.
The solution’s temperature must be raised to
make the solution’s vapor pressure equal to the
atmospheric pressure.
The amount that the temperature is elevated
is determined by the number of moles of
solute dissolved in the solution.
33
Boiling Point Elevation

Boiling point elevation relationship is:
Tb  K b m
where : Tb  boiling point elevation
m  molal concentrat ion of solution
K b  molal boiling point elevation constant
for the solvent
34
Boiling Point Elevation

Example 14-4: What is the normal boiling
point of a 2.50 m glucose, C6H12O6, solution?
Tb  K b m
Tb  (0.512 0 C/m)( 2.50m)
Tb  1.280 C
Boiling Point of the solution = 100.0 0 C + 1.280 C = 101.280 C
35
Boiling-Point Elevation
The addition of a nonvolatile solute lowers the vapor
pressure of the solution.
At any given temperature,
the vapor
pressure
of the
solution
is lower
than that
of the
pure
liquid
36
The increase in boiling point relative to that of the
pure solvent, Tb, is directly proportional to the
number of solute particles per mole of solvent
molecules.
Molality expresses the number of moles of solute per
1000 g of solvent, which represents a fixed number
of moles of solvent
Tb  K b m
B.Point (°C)
Solvent
Water, H2O
Benzen, C6H6
Ethanol, C2H6O
Carbon tetrachloride,
CCl4
Chloroform,
CHCl3
100.0
80.1
78.4
76.8
61.2
Kb
(°C/m)
0.52
2.53
1.22
5.02
3.63
Freezing P.
(°C)
Kf (°C/m)
0.00
5.5
-114.0
-22
-63.5
1.86
5.12
1.99
29.8
4.68
37
Automotive antifreeze consists of ethylene glycol,
C2H6O2, a nonvolatile nonelectrolyte. Calculate the
boiling point of a 25.0 mass percent solution of
ethylene glycol in water.
38
Freezing Point Depression


Addition of a nonvolatile solute to a solution
lowers the freezing point of the solution
relative to the pure solvent.
See table for a compilation of boiling point
and freezing point elevation constants.
39
Freezing Point Depression

Relationship for freezing point depression is:
Tf  K f m
where: Tf  freezing point depression of solvent
m  molal concentration of soltuion
K f  freezing point depression constant for solvent
40
Freezing Point Depression

Notice the similarity of the two relationships
for freezing point depression and boiling point
elevation.
Tf  K f m vs. Tb  K b m

Fundamentally, freezing point depression and boiling
point elevation are the same phenomenon.


The only differences are the size of the effect which is
reflected in the sizes of the constants, Kf & Kb.
This is easily seen on a phase diagram for a solution.
41
Freezing Point Depression
42
Freezing Point Depression

Example 14-5: Calculate the freezing point of
a 2.50 m aqueous glucose solution.
Tf  K f m
Tf  (1.86 0 C/m)( 2.50m)
Tf  4.650 C
Freezing Point of solution = 0.00 0 C - 4.650 C = - 4.650 C
43
Freezing Point Depression

Example : Calculate the freezing point of a
solution that contains 8.50 g of benzoic acid
(C6H5COOH, MW = 122) in 75.0 g of
benzene, C6H6.
You do it!
44
Freezing Point Depression
1. Calculate molality!
? mol C 6 H 5 COOH 8.50 g C 6 H 5 COOH


kg C 6 H 6
0.0750 kg C 6 H 6
1 mol C 6 H 5 COOH
 0.929m
122 g C 6 H 5 COOH
2. Calculate the depression for this solution.
Tf  K f m
Tf  (5.12 0 C/m)( 0.929m)  4.76 0 C
F.P. = 5.480 C - 4.76 0 C = 0.72 0 C
45
Determination of Molecular Weight by
Freezing Point Depression
The size of the freezing point depression
depends on two things:

1.
2.

The size of the Kf for a given solvent, which are
well known.
And the molal concentration of the solution which
depends on the number of moles of solute and
the kg of solvent.
If Kf and kg of solvent are known, as is
often the case in an experiment, then we
can determine # of moles of solute and use
it to determine the molecular weight.
46
Determination of Molecular Weight by
Freezing Point Depression

Example : A 37.0 g sample of a new covalent
compound, a nonelectrolyte, was dissolved in
2.00 x 102 g of water. The resulting solution
froze at -5.58oC. What is the molecular
weight of the compound?
47
Determination of Molecular Weight by
Freezing Point Depression
Tf  K f m thus the
Tf 5.580 C
m

 3.00m
0
K f 1.86 C
In this problem there are
200 mL  0.200 kg of water.
? mol compound in 0.200 kg H 2 O = 3.00 m  0.200 kg
 0.600 mol compound
37 g
Thus the molar mass is
 61.7 g/mol
0.600 mol
48
Osmotic Pressure
Osmosis is the net flow of a solvent
between two solutions separated by a
semipermeable membrane.


The solvent passes from the lower concentration
solution into the higher concentration solution.
Examples of semipermeable membranes
include:

1.
2.
3.
cellophane and saran wrap
skin
cell membranes
49
Osmotic Pressure
semipermeable membrane
sugar dissolved
H2O
in water
H2O
H2O
H2O
H2O
H2O
H2O
net2solvent
flow
O
50
Osmotic Pressure

Osmosis is a rate controlled phenomenon.


The solvent is passing from the dilute solution into
the concentrated solution at a faster rate than in
opposite direction, i.e. establishing an equilibrium.
The osmotic pressure is the pressure exerted
by a column of the solvent in an osmosis
experiment.
  MRT
where:  = osmotic pressure in atm
M = molar concentration of solution
L atm
mol K
T = absolute temperature
R = 0.0821
51
Osmotic Pressure
 For very dilute aqueous solutions, molarity
and molality are nearly equal.
 Mm
  mRT
for dilute aqueous solutions only
52
Osmotic Pressure
Osmotic pressures can be very large.

For example, a 1 M sugar solution has an osmotic
pressure of 22.4 atm or 330 p.s.i.

Since this is a large effect, the osmotic
pressure measurements can be used to
determine the molar masses of very large
molecules such as:

Polymers
Biomolecules like
1.
2.


proteins
ribonucleotides
53
Osmotic Pressure

Example : A 1.00 g sample of a biological
material was dissolved in enough water to give
1.00 x 102 mL of solution. The osmotic pressure
of the solution was 2.80 torr at 25oC. Calculate
the molarity and approximate molecular weight
of the material.
You do it!
54
Osmotic Pressure
  MRT  M 

RT
1 atm
? atm = 2.80 torr 
 0.00368 atm = 
760 torr
0.00368 atm
4
M =

150
.

10
M
L atm
0.0821 mol K 298 K
55
Osmotic Pressure
  MRT  M 

RT
1 atm
? atm = 2.80 torr 
 0.00368 atm = 
760 torr
0.00368 atm
4
M =

150
.

10
M
L atm
298 K 
0.0821 mol
K 
?g
1.00 g
1L
4g



6
.
67

10
mol
4
mol 0.100 L 150
.  10 M
typical of small proteins
56
Application of Osmotic Pressure
1. Water Purification by Reverse Osmosis
 If we apply enough external pressure to an
osmotic system to overcome the osmotic
pressure, the semipermeable membrane
becomes an efficient filter for salt and other
dissolved solutes.


Ft. Myers, FL gets it drinking water from the Gulf
of Mexico using reverse osmosis.
Dialysis is another example of this phenomenon.
57




2) Isotonic solution: In the living cells, the
osmotic pressure of solution is equal to
the osmotic pressure of the cell.
e.g: NaCI (0.9%) has the same osmotic
pressure as blood.
3) Hypertonic solution: A solution of
higher osmotic pressure. In this solution
red blood cells shrink. The cells are called
plasmolysed.
4) Hypotonic solution: A solution of lower
osmotic pressure. In this solution red
blood cells swells up and burst. The cell
is said to be haemolysed
58
End of Chapter 2

Human Beings are solution chemistry in
action!
59