1

IDEAL SOLUTION To summarize then, for ideal solutions at constant

• Solution is ideal when molecules of the T and P,
components are so similar to one another mixV  0 mix S  R  ni ln xi
• replacing molecules of one species with
mix H  0 mix G  RT  ni ln xi
molecules of another species will not change
mix U  0
the spatial structure or molecular interaction
energy Example:
• Concept of ideal solution will help us
At 200C and 1 atm, density of benzene is
understand real solutions 0.879 g/ml and that of toluene is 0.8668 g/ml.
• Closest resemblance occurs for isotopic Find the density of a solution of 33.33 g of
species benzene and 33.33 g of toluene at this
Example: CHCl3 and CHBr2 condition. Assume ideal solution.
• Also with liquids of similar structures and size.
Raoult’s Law
Example: n-C7H6 and n-C8H18,
C2H5Cl and C2H5Br  Francois Marie Raoult observed that in an
C(CH3)4 and Si(CH3)4 ideal solution, the vapor pressure of a
component is proportional to its mole fraction
Thermodynamics of Ideal Solution
in the liquid solution
• On mixing the pure constituents, the change in
• PA  x A PA0
volume is zero.
where: xA = mole fraction of component A in
 Gmix 
   0  Vmix the liquid solution
 P T
PA0 =vapor pressure of A when pure
• This implies that an ideal solution is formed
 Total vapor pressure
without evolution or absorption of heat
PT  PA  PB  x A PA0  xB PB0
 Gmix 

 T

 0
Hmix
 Hmix  0

 x A PA0  1  x A  PB0  PB0  x A PA0  PB0 
 T  T2 This is linear in xA
 
 P
• Smix is a function of concentration and
quantity of constituents, not of their identity
Smix  nA R ln x A  nB R ln xB
 nR  x A ln x A  xB ln xB 

In most cases, Smix>0, but for some like HCl

(g) in water, Smix<0

 From: Hmix  Umix  P Vmix ,

0  Umix  P  0 
Let A = solute, B = solvent
Umix = 0
If A is nonvolatile, then
PT  PB  xB PB0  PB0
There is a lowering in vapor pressure
 2

P  PB0  xB PB0  PB0 1  xB  Calculate the mole fraction of benzene in the

vapor phase when vapor is in equilibrium with
 x A PB0
the liquid.
 In the vapor: 3. If the vapor pressure of pure liquids A and B
y A  mole fraction of A in the vapor
are 300 mm and 800 mm at 75 0C respectively,
PA x A PA0 calculate the composition of the mixture such
= 

PT PB0  PA0  PB0 x A  that it boils at 75 0C. Also find the composition
Solving for x A and substitute in the of the vapor phase.
4. At 300 K, the vapor pressure of an ideal
equation for PT
solution containing one mole of A and 3 moles
PA0 PB0
PT  of B is 550 mm Hg. If one mole of B is added

PA0  PB0  PA0 y A  to this solution at the same temperature, the
vapor pressure of the solution increases by 10
Taking reciprocals and substituting the
mm Hg. Calculate the vapor pressure of A and
expression for y A ,
y y
B in their pure states.
1
 A  B 5. A solution containing 30 g of a non-volatile
Pt 0
PA PB0
solute exactly in 90 g of water has a vapor
If the liquid line L and the vapor curve V
pressure of 2.8 kPa at 298 K. 18 g more of
are drawn on the same coordinate system
water is added to the solution and the new
the vapor curve is below the liquid line.
vapor pressure becomes 2.9 kPa at 298 K.
Calculate
a. molecular mass of the solute
b. vapor pressure of water at 298 K
6. The normal BP of methanol is 64.7 0C. A
solution containing a nonvolatile solute
dissolved in methanol has a vapor pressure of
710 torr at 64.7 0C. What is the mole fraction
of methanol in this solution?
L = liquid curve or liquidus 7. Calculate the mass of non-volatile solute
V = vapor curve (MW = 40) which should be dissolved in 114
g of octane to reduce its vapor pressure to
Exercise 80%.
1. Heptane and octane form ideal solution. At 8. Vapor pressure of C6H6 and C7H8 mixture at
373 K, the vapor pressures of the two liquid 50 0C are given by P = 179 XB +92 (in mm Hg),
component are 105.2 kPa and 46.8 kPa where XB is mole fraction of benzene. Find
respectively. What will be the vapor pressure a. vapor pressure of the pure liquids
of a mixture of 25 g of heptane and 35 g of b. vapor pressure of liquid mixture
octane? obtained by mixing 936 g C6H6 and 736
2. The vapor pressure of benzene and toluene g C7H8
at 200C are 75 mm Hg and 22 mm Hg c. If the vapors are removed and
respectively. 23.4 g of benzene and 64.4 g of condensed into liquid and again
toluene are mixed to form an ideal solution. brought to the temperature 50 0C, what
 3

would be the mole fraction of benzene 2. The vapor pressures of ethanol and
in the vapor state? methanol are 44.5 and 88.7 mm Hg
9. One mole of benzene (component A) is respectively. An ideal solution is formed at
mixed with 2 moles of toluene (component the same temperature by mixing 60 g of
B). At 60 0C the vapor pressures of benzene ethanol and 40 g of methanol. Calculate the
and toluene are 51.3 and 18.5 kPa, total vapor pressure of solution and mole
respectively. Draw the vapor pressure- fraction of methanol in the vapor phase.
composition diagram for the system. 3. What mass of solute (MW = 60) is required
10. The vapor pressure of 2 pure liquids A and B to dissolve in 180 g of water to reduce the
that form an ideal solution are 300 and 800 vapor pressure to 4/5 that of pure water?
torr respectively at temperature T. A mixture 4. Find the molality of a solution containing a
of the vapors of A and B for which the mole non-volatile solute if the vapor pressure is 2 %
fraction of A is 0.25 is slowly compressed at below the vapor pressure of pure water.
temperature T. Calculate 5. Solutions of 2 volatile liquids A and B obey
a. the composition of the first drop of Raoult’s law. At a certain temperature, it is
the condensate found that when the total pressure above a
b. the total vapor pressure when this given solution is 400 mm Hg, the mole fraction
drop is formed of A in the vapor is 0.45 and in the liquid, 0.65.
c. composition of the solution whose What are the vapor pressures of the 2 pure
normal BP is T liquids at the given temperature?
d. the pressure when only the last 6. Calculate the vapor pressure lowering of a
bubble of vapor remains 0.10 m aqueous solution of non-volatile non-
e. composition of the last bubble electrolyte at 75 0C.
11. The vapor pressure of pure acetone at 55.1 7. A mixture of 2 immiscible liquids nitrobenzene
0
C is 632.8 mm and that of CHCl3 is 741.8 mm. and water boiling at 99 0C has a partial vapor
Assuming that they form ideal solution over pressure of water equal to 733 mm and that of
the entire range of composition, plot total nitrobenzene 27 mm. Calculate the ratio of
pressure versus mole fraction of acetone. the weights of nitrobenzene to that of water
The actual data observed for different in the distillate.
composition of mixture is given below:

xacetone 0 0.118 0.236 0.36 0.508 0.582 0.645 0.721

Pacetone 0 54.9 110.1 202.4 322.7 405.9 454.1 521.1
PChloroforml 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7

Problem Set RAOULT’S LAW AS A SPECIAL CASE OF HENRY’S

LAW
1. The vapor pressure of water is 12.3 kPa at 300
According to Raoult’s law, the vapor pressure
K. Calculate the vapor pressure of 1 molal
of a volatile component of a solution is
aqueous solution the solute of which is
pA  x A PA0
nonvolatile and non-electrolyte
For a solution of a gas in a liquid, the gaseous
component is taken to be a volatile
 4

component. Its solubility is given by Henry’s

law
P  KH x
Comparing the two equations, gives the
conclusion that the pressure of a volatile
component is directly proportional to its mol
fraction. The expressions become identical  If A-A and B-B attractions
whe KH  PA0 . This shows that Raoult’s law is exceed A-B attractions,
 vapor pressure higher
a special case of Henry’s law. Even for liquid than that predicted by Raoult’s Law.
solutions, one component which is volatile, PA  x A PA0 ; PB  xB PB0
may sometime obey Henry’s law over certain  Hmix > 0
range of concentrations. In practice, in dilute  Vmix > 0
solutions, if the solute obeys Henry’s law for  at some intermediate composition of the
a given range of concentrations, the solvent solution, the vapor pressure is maximum
obeys Raoult’s law for same range of and the BP is minimum. This composition
concentrations. The reverse however is not refers to minimum-boiling azeotrope
necessarily true.  Example:
CCl4  C6 H5CH3  CH3 2 CO  C2 H5OH
LIQUID-VAPOR PHASE DIAGRAMS FOR H2 O  CH3OH C6 H6   CH3 2 CO
COMPLETELY MISCIBLE LIQUIDS H2 O  C2 H5OH CCl4  CHCl3
 CH3 2 CO  CS2
If solution is ideal, the liquidus and vapor curve
are as shown
Negative Deviation from Raoult’s law

Lever Rule
 If A-A and B-B attractions are less than A-
nv x x B attractions,
 B  vapor pressure lower than that predicted
nL x  yB
by Raoult’s Law.
La
 PA  x A PA0 ; PB  xB PB0
aV
 Hmix < 0
 Vmix < 0
 at some intermediate composition of the
Line connecting two points on the graph is solution, the vapor pressure is minimum
and the BP is maximum. This
called a tie line.
composition refers to maximum-boiling
Positive Deviations from Raoult’s law azeotrope.
 5

 Example: Liquid with minimum boiling composition

H2 O  HCl CHCl3   CH3 2 CO is called an azeotrope. At this
CHCl3  C6 H6 CHCl3   C2 H5 2 O composition, vapor and liquid have the
H2 O  HNO3 CH3 2 CO  C6 H5NH2 same composition
Example:
Solutions of Two completely Miscible Liquids methylal and CS2
There are 3 types: C6H6 and cyclohexane,
C6H6 and C2H5OH,
CCl4 and C2H5OH,
H2O and C2H5OH,
CS2 and CH3COOCH3

Type 3: System exhibits a minimum vapor

pressure or maximum boiling point.

Type 1. Boiling point is between those of the

pure components

 This type exhibits a maximum boiling

azeotrope
 Examples are:
Acetone and CHCl3 CH3OCH3 and HCl
H2O and HCOOH H2O and HNO3
H2O and HCl H2O and HBr
Example:
cyclohexane and CCl4, CCl4 and C6H6 Distillation of Binary system
C6H6 and C6H5CH3, H2O and CH3OH
Type 1.
Type 2. System exhibits a maximum vapor
pressure or a minimum boiling point

Consider a solution corresponding to composition

a.
 6

No boiling will start until temperature Ta is Example of this type is water (A) – ethanol (B)
reached. The composition of the vapor phase system. if a solution of composition between A
at this stage will a’ that is, it is richer in the an C such as a, is distilled, the vapor coming off
component B. The residue then become will have the composition a’ and will be richer in
richer in A shifting the composition towards B and the residue will become richer in B, and the
A, say equal to b. If this liquid mixture is residue will become richer in A and the
heated, it will boil only when the composition of the residue will shift towards A till
temperature becomes equal to Tb. The vapor eventually a residue of pure A will be obtained.
will have the composition b’ which is richer in
The liquid obtained on condensing the vapors
B, and consequently the composition of the
(corresponding to composition a’), if distilled will
residue will be further enriched in A. Thus if
give vapor richer in B. If the condensation of
the process of heating the residue is
vapors and the distillation of the liquid is
continued, the BP of the solution will rise
obtained, ultimately the vapors of composition C
from the initial BP Ta towards the BP TA. of
will be obtained. if these vapors are condensed
the pure liquid A. Moreover, every time, the
and the solution distilled, the distillate obtained
residue becomes richer in A than the original
will have the same composition as the solution.
solution. this means that if the process is
Hence no further separation is possible by
continued for sufficiently long time, a final
distillation. Thus in case of solutions of type ii, a
residue of pure A can be obtained.
solution of composition between A and C on
If the vapors obtained in the first stage are fractional distillation gives residue of pure A and
condensed we shall get a liquid mixture a final distillate of composition C. No pure B can
corresponding to composition a’. If this liquid be recovered.
mixture is distilled, it will boil when when the Consider now the distillation of a solution having
temperature becomes Ta’. The composition a composition between C and B, say
of the vapor coming off will correspond to corresponding to the point b. The vapor coming
the point a’’, that is the vapors become richer off will be richer in A and so the residue will be
in B than the original solution this means richer in B. Hence on repeated distillations
that if the process of condensing the vapor ultimately pure B can be obtained from the
and redistilling the liquid mixture is residue and the final distillate will be of
continued, ultimately a distillate of pure B is composition C. No pure A can be recovered.
obtained.
Type iii
For type 1 solution therefore, a complete
separation of the components is possible by
distillation. The less volatile being left as the
residue and the more volatile component
being obtained as the distillate

Type ii.

The behavior is analogous to that of type ii with

the exception that the residue tend towards the
constant boiling mixture corresponding to point D
whereas distillates tend towards the pure
constituents. compositions between A and D say
 7

a, the vapor coming off are richer in A hence the

residue is richer than B. The composition of the
residue shifts towards D and ultimately becomes
equal to that of D. The composition of the vapors
shifts towards A and finally a distillate of pure A is
obtained .

Thus for a mixture having composition between A

and D, distillate of pure A is obtained and the
residue has a composition corresponding to point
D. similarly, for a mixture having composition
between D and B, ultimately distillate of pure B
and residue of composition corresponding to D is Plot the shown BP data for benzene-ethanol
obtained. solutions at 1.013 bar, and
a. Find the azeotropic composition.
Exercise
b. Find the range of mol fraction of C6H6
1. Given the phase diagram below:
for which pure C6H6 could be obtained
by fractional distillation at 1.013 bar.

For a liquid mixture consisting of 3 mol of

A and 7 mol of B.
a. At what T will the mixture begin to
boil? What is composition of the vapor?
b. If the distillation is continued until BP
is raised by 5 C0, what is the composition
of liquid left in the still ?

2. Given the data below:

3. Consider the phase diagram of 2 soluble
o
Mole fraction of C6H6
BP, C liquids, as shown. Find the number of
In liquid In vapor theoretical plates necessary to separate the
78 0 0 pure form of B from a solution of composition
75 0.04 0.18 x1 .

70 0.21 0.42

70 0.86 0.66

75 0.96 0.83

80 1.00 1.0
 8

2. The following data is given about the liquid-

vapor phase diagram i.e., the BP of various
composition of liquids A and B

Tb, 0C 53.4 54 55.5 58.6 60 62.8

XA(l), 0 3.5 12 30.65 45.1 61.1
XA(v) 0 2.5 7.9 25.42 34.8 59.5
Tb, 0C 63 62.2 61.2 60 59.2
XA(l), 70 82.3 86.7 95.1 100
XA(v) 77.5 84.1 91.5 97 100

a. Draw the liquid-vapor phase diagram as Example is phenol-water system.

BP vs mole fraction. Supposing phenol and water are mixed in
b. Find the azeotropic temperature and equal proportions, two layers are produced:
azeotropic composition one is a solution of phenol in water and the
c. Take a mixture containing liquid as 0.35 other a solution of water in phenol. At any
and heat up to 59.5 0C in a closed given temperature, the composition of the
system at 1 atm. What will be the layers is fixed. If the system is warmed the
composition of the liquid phase and that amount of water in phenol layer increases as
of the vapor phase? well as the amount of phenol in the water
d. If 60 g of A (MW = 48 g/mol) and 120 g of layer increases. Ultimately at a particular
B (MW = 80) are heated up to 62 0C in a temperature, the composition of both layers
closed vessel at a atm, calculate the becomes the same and thus the two become
amount of liquid phase and vapor phase. completely miscible. Experiments showed
that the minimum temperature at which
Partially Miscible Liquids phenol and water become completely
miscible is 66 0C and the composition is 34 %
 Miscible within certain range of phenol by mass. Thus phenol-water system
temperature has an upper consolute or critical
 Critical Solution Temperature or temperature. This is shown in the plot below
Consolute temperature
Temperature at which the 2 liquids
which are otherwise partially miscible
at ordinary temperature, become
completely miscible. There are 3 types.

Type 1. System exhibits an upper consolute

temperature

Type 11: System exhibits a lower consolute

temperature

Example: Triethylamine-water system

 9

mixed at 25 0C and at this temperature the

system forms two phases. if the mass of B in
phase 1 is 15 % and that of B in phase 11 is 80
%, calculate the masses of each phase in
equilibrium

The mutual solubility of the two liquids

increases with decrease of temperature, thus
on cooling the system ultimately a stage is
reached when the two liquids become
completely miscible. This temperature is
2. Water and phenol are partially miscible at
called the lower consolute temperature and
500C. When these two liquids are mixed at
is 18.5 0C for triethylamine-water system.
500C and 1atm, at equilibrium one phase is 89
This means that below this temperature
% water by mass and the other is 37.5 % water
triethylamine and water are completely
by mass. If 6 grams of phenol and 4 grams of
miscible in all proportions.
water are mixed at 500C and 1 atm, find the
mass of water and the mass of phenol in each
phase at equilibrium.
3. Hexane and perflurohexane show partial
Type 111. System exhibits both lower and upper miscibility below 22.7 0C. The critical
consolute temperatures concentration at the upper critical
temperature is x = 0.355, where x is the mole
fraction of hecane. At 22 0C the 2 solutions in
equilibrium have x = 0.2 and x = 0.48,
repectively, and at 21.5 0C, mole fractions are
0.22 and 0.51 respectively. Sketch the phase
diagram. Describe the phase changes that
occur when perflurohexane is added to a fixed
amount of hexane at
a. 23 0C b. 22 0C

Immiscible Liquids
 Vapor pressure of each component
independent of the other

Mutual solubility of the two liquids increases Pt  P1  P2  P10  P20

with increase of temperature as well as with  mixture boils at a lower temperature
decrease of temperature. The two liquids
become completely solubility above a Let N1*  mole fraction of component 1 in
particular temperature as well as below the vapor phase
another temperature. That is the system has
P10  N1* Pt P20  N2* Pt
an upper critical temperature and a lower
critical temperatures. P10 N1* n1
   a constant
P20 N2* n2
Exercise
m1  MW2 
1. Consider the following liquid-liquid phase   
MW1  m2 
diagram. In this, 25 g of A and 60 g of B are
m1 P10  MW1 
  
m2 P20  MW2 
 10

Nernst’ Equation
If a 3 rd substance is added to a 2-phase
system of immiscible liquids, it will distribute
This can be used to determine the MW of a itself between the two solvents until at
liquid from that of a known one equilibrium, the ratio of the activities of the
substance in the two layers is constant at a
given temperature.
Steam Distillation
 used in the lab or in the industry for the aA
K
purification of organic liquids. aB
 process is generally applied under the K = distribution or partition coefficient of
following conditions: solute between 2 solvents
a. organic liquid to be purified should be
When solutions are dilute, a = molarity,
immiscible with water
b. it should have high MW CA
C  K
c. it should have a high vapor pressure at CB
about 1000C Note: K depends on the nature of solute and
d. impurities present should be solvents, temperature, and manner
nonvolatile Nernst Law is written
 Set up  CA CB 
 whether as or as 
 CB CA 

Example
To 0.1 L of water containing 0.4 M SO2, 0.1 L of
CHCl3 was added. Calculate the moles of SO2
present in the two immiscible solvents water
and chloroform after equilibrium is reached.
Exercise For SO2,
1. A mixture of water and aniline boils at a CH2O
 0.98
temperature of 98.5 0C at pressure of 760 mm. CCHCl3
The vapor pressure of water at this
Solution:
temperature is 717 mm. Find the composition
(as % mass) of the distillate.
2. An organic liquid was subjected to steam
distillation. The liquid in the flask boiled at
900C. The external pressure (atmospheric)
was found to be 734.4 mm Hg. The vapor
pressure of water at this temperature is 526
mm. In the distillate, the ratio of the masses
of liquid to water is found to be 2.47.
Calculate the MW of the liquid.
3. Naphthalene may be steam distilled at 99.3 0C
under atmospheric pressure. What weight of
steam will be required to carry 2 kg of
naphthalene into the distillate at atmospheric
pressure?
 11

Initially, mass of SO2 in water

 mol 
=  0.4   0.1 L   0.04 mol
 L 
At equilibrium:
Let x  mole of SO2 in water
0.04  x  mole of SO2 in CHCl3
x
0.1  0.98
0.04  x After second extraction ( n = 2)
0.1
x  0.0198 mole in water, and WA 2
CA V W  VB 
0.0202 mole in CHCl3 k   A  A2  
CB WB 2 WB 2  VA 
VB
Exercise
KVAWB 2 KVA
WA 2   WA  WA2 
1. 50 mL of water and 25 mL of CCl4 were taken VB VB
in a cylinder. 4 g of I2 was added. calculate KVA  KVA 
  W  WA 2 
the amount of iodine present in each layer by VB  VB  KVA 
using Nernst law. Nernst coefficient
After the n the extraction
CCCl4
 86  KVA 
n
CH2O WAn    W
 VB  KVA 
2. In a cylinder containing 50 g water and 70 g  kVA 
W  W  Wn = W   W
benzene, some HgCl2 was added. Calculate  kVA  VB 
C   kV  
n
the Nernst constant water if after = W 1   A
 
C benzene   kVA  VB  
 
equilibrium it is found that water contains  kVA 
n
W
Fraction Extracted,  1  
0.204 g and benzene contains 0.0217 g of W  kVA  VB 
HgCl2. Density of benzene is 1.5 kg/L
3. The vapor pressure of the immiscible Example:
liquid system diethylaniline-water is The distribution coefficient of lactic acid
1.013 atm at 99.4 0C. The vapor pressure between water and chloroform
of water at this temperature is 99.2 kPa. CH2O 1
How many grams of steam are necessary  at 25 0C.
to distill 100 g of diethylamine? CCHCl3 0.023
200 mL of chloroform contains 1 M solution of
Solvent Extraction lactic acid. Calculate the amount of lactic acid
withdrawn when 100 mL of water is shake with
solution
a. once
b. twice by taking 50 mL each time
c. 4 times by taking 25 mL each tim

Problem Set
WA
CA V W V  KVAWB
k  A  A  B   WA 
CB WB WB  VA  VB
VB
KVA KVAWA KVAW
WA  W  WA   WA  
VB VB VB
 12

1. Given the following data COLLIGATIVE PROPERTIES OF SOLUTIONS

Depend on the quantity of solute molecules.
BP, 0C 100 92 89.3 88.2 87.8
xprop 0 2 6 20 43.2
Example:
yprop 0 21.6 35.1 39.2 43.2 1. Relative Lowering in Vapor Pressure
BP, 0C 88.3 90.5 97.3 2. Freezing point depression
xprop 60 80 100 3. Melting point elevation
yprop 49.2 64.1 100 4. Osmotic Pressure

Plot the graph, and calculate the mole fraction

of n-propanol in the first drop of distillate
when the following solutions are distilled with
a simple distilling flask that gives one
theoretical plate.
a. 87 g of n-propanol and 211 g water
b. 50 g of n-propanol and 5.02 g water

2. At 300C a mixture of phenol and water is made

up containing 60 % water. The mixture splits
into 2 layers, the phenol layer containing 70 %
phenol and the water layer containing 92 %
water. All % are by mass. Calculate the
Relative Lowering in Vapor Pressure
relative masses of the 2 layers.
P10  Psoln
3. The boiling point of the immiscible liquid  x2
P10
system naphthalene-water is 98 0C under a
pressure of 97.7 kPa. The vapor pressure of Freezing Point Depression
water at this temperature is 94.3 kPa.
For the solvent,
Calculate the mass % of naphthalene in the
distillate. Gfus
ln x1  
RT
Differentiate with respect to T:
 Gfus 
d 
4. A totally immiscible liquid system composed of d  ln x1  1  T  1  H 
    
water and an organic liquid boils at 90 0C dT R dT R  T2 
when the barometer reads 734 mm Hg. The H  dT 
distillate contains 73 % by weight, organic d  ln x1    
R T2 
liquid. What is the molecular weight and
Integrate:
vapor pressure at 90 0C of the organic liquid?
H  dT 
x11d ln x1   Tf0
x1 T
At 90 0C, vapor pressure of water is 525.8 torr  
R T2 

5. At 20 0C, SO2 was permitted to distribute itself H  1 1 

ln x1     
between 200 cc of CHCl3 and 75 cc of water. R  T Tf0 
When equilibrium was established, the Recall: ln x1  ln 1  x2 
chloroform layer contained 0.14 mole of SO2 Power series expansion gives ln x1   x2
and the water layer 0.05 mole. What is the
distribution coefficient of SO2 between water
and chloroform at 20 0C?
 13

1H 1  Similarly it can be shown that:

Thus ln x1  x2     
T T 0
R  Hvap  1 1 
 f  ln x1     or
R  T Tb0 
H  Tf  T  
0
  
R  T Tf0  1 1 R ln x  RT02 
  Tb    x2
Note: Tf0  T  Tf T Tb 0 Hvap  Hvap 
 
and since Tf is small, T  Tf0 For dilute solutions, Tb  kb m
Hence:
where: kb  molal boiling point or

x2 

Hfus  Tf



 R Tf
0 2
  
 ebullioscopic constant
   Tf    x2
R  0 2
T     Hfus 
 f   
   
2
R Tb0 MW1 R Tb0 MW1
where: Tf0 and Hfus are properties Kb = 
of component 1 (the solvent) H vap Svap
while x 2 is the mole fraction Note:
of component 2 (the solute) kb is high for large molecular weights
n2
Recall: x2  Colligative Constants of some Solvents
n1  n2
For dilute solutions, Solvent BP, Kb, FP, 0C Kf ,
0
n n n C C0/m 0
C/m
x2  2 = 2  2 MW1 Acetic acid 118.3 3.07 16.6 3.57
n1 W1 W1
MW1 Benzene 80.2 2.53 5.45 5.07
Camphor - - 178.4 37.7
If m  molality of solute, x2  m  MW1  CS2 46.2 2.34 - 111.5 3.83

 
 0 2 CCl4 76.5 5.03 - 23 30.00
 R Tf MW1 
Chloroform 61.2 3.63 - 63.5 4.70
Tf    m  kf m
 H fus  Cyclohexane 80.7 2.69 6.5 20
  Diethyl ether 34.5 2.02 -116.2 1.79
   
2 - 117.3
R Tf0 MW1 R Tf0 MW1 Ethanol 78.5 1.22 1.99
kf   Naphthalene - - 80.2 6.8
H fus S fus Water 100 0.512 0 1.86
Note:
 kf is called the molal freezing point or
cryoscopic constant
 kf is large for solvents of high molar mass
Osmotic Pressure
 Freezing point depression is large if the
Osmosis:
solvent has large molecular weight.
the movement of a solvent through a semi-
permeable membrane from low solute
From the definition of molality,
W2
concentration to high solute concentration
MW2 W2
m  Tf  mkf  kf
W1 W1  MW2 
Solve for the molar mass of the solute:
kf  W2 
MW2   
Tf  W1 
Note: Freezing point depression can be used to
determine the molar mass of solutes

Boiling Point Elevation

 14

penetration of solvent into it through the

semi-permeable membrane.
• If pressure greater than the osmotic
pressure is applied on the solution, then
the solvent particles start passing through
semi-permeable membrane towards the
region of pure solvent. This phenomenon
is referred to as reverse osmosis
• Two solutions having same osmotic
pressure are said to be isotonic.
• Among two solutions having different
There is movement in both directions across a
osmotic pressures, the solution with lower
semipermeable membrane. As solvent moves
osmotic pressure is said to be hypotonic
across the membrane, the fluid levels in the
while the one with the larger osmotic
arms becomes uneven. Eventually the
pressure is said to be hypertonic.
pressure difference between the arms stops.
Effect on Cells
The excess hydrostatic pressure which builds
up as a result of osmosis is called osmotic
pressure

Van’t Hoff’s Equation for osmotic pressure:

n2
  RT  CRT
V
where: C = molarity of solution
R = 0.08206 L atm/mol K
Note: This equation is of similar form as the
ideal gas equation.
Colligative properties can be employed to Exercises
determine the molecular mass of
substances. Most practical among them 1. A solution containing 0.730 g of camphor (MW
is osmotic pressure. = 152) in 36.8 g of acetone (BP = 56.30 0C)
boils at 56.55 0C. Calculate the MW of the
Osmosis vs Diffusion
Osmosis Diffusion
Osmosis Diffusion
There is a flow of There is a flow of both
Solvent flows from Solution flows from
solvent into the solute and solvent
the solution of higher solute
solution through a and no semi-
lower solute concentration to
semi-permeable permeable membrane
concentration to lower solute
membrane is required
solution of higher concentration until an
unknown compound.
solute equilibrium in
concentration concentration is 2. The BP of a solution containing 0.5 g of solute
achieved. B (MW = 0.128 kg/mol) per 50 g of solvent is
raised by 0.4 K while 0.6 g of another solute C
Important features raises the BP of 50 g of same solvent by 0.6 K.
• Osmotic pressure may also be defined as Calculate the MW of C.
the excess pressure that must be applied
to the solution side to prevent the
 15

3. Calculate the BP of 1 M aqueous solution of a 20 mm of Hg at 27 0C. Calculate the MW of

solute (MW = 74.5). The density of the protein.
solution is 1.04 g/mL) 13. The osmotic pressure of blood at 37 0C is 8.21
4. Two elements A and B form compound having atm. How much glucose should be used per
formulas AB2 and AB4. when dissolved in 20 g liter for an intravenous injection so that it is
of benzene, 1.0 g of AB2 lowers the FP by 2.3 K isotonic with blood?
whereas 1.0 g of AB4 lowers the FP by 1.3 K. 14. Two solutes X and Y in equal amounts are
The molal cryoscopic constant for benzene is dissolved separately in 50 g of solvent C. If X
5.1 K kg/mol. Calculate the atomic masses of molecule is heavier than Y, which solution
A and B. will show greater lowering of vapor pressure?
5. A solution containing 18 g of non-volatile 15. An organic compound contains 68 % C, 10 %
solute in 200 g of water freezes at 272.07 K. H and the rest, oxygen. A solution of 0.032 g
Calculate the MW of the solute. of the compound in 0.722 g of cyclohexane
has a freezing point of 6.39 C 0 less than that
6. A solution containing 25.6 g of sulfur dissolved
of the solvent. Calculate the molar mass and
in 000 g of naphthalene gave a FP lowering of molecular formula of the compound.
0.680 Calculate the molecular formula of 16. At 10 0C, the osmotic pressure of urea
sulfur. For naphthalene, Kf = 6.8 K kg/mol. solution is 500 mm Hg. The solution is
7. A radiator of motor vehicle was filled with 8 L diluted and the temperature is raised to 25
0
of water to which 2 L of methanol (density = C, when the osmotic pressure is found to be
0.8 g/ml) were added. What is the lowest 105.3 mm. Determine extent of dilution.
temperature at which the vehicle can be
Problem Set
parked without a danger of getting water in
1. A solution of a nonvolatile solute in water has
the radiator to freeze?
a BP of 375.3 K. Calculate its vapor pressure of
8. How many grams of solute (MW = 342) should water above this solution at 338 K. The vapor
be dissolved in 500 g of water so as to get a pressure of pure water at this temperature is
solution having a difference of 105 C0 between 0.2467 atm.
FP and BP? 2. The vapor pressure of water at 293 K is 2338
9. Two aqueous solutions of A and B freeze at Pa and the vapor pressure of an aqueous
solution is 2295.8 Pa. If the density of the
the same temperature. A is a solution of 7.5 g
solution is 1010 kg/m3 at 313 K, calculate the
of urea in 100 g of water whereas B is a
osmotic pressure at this temperature. MW of
solution of X in 100 g of water. What is the solute is 60.
MW of X? 3. Find the molality of a solution containing a
10. Calculate the osmotic pressure of a solution non-volatile solute if the vapor pressure is 2 %
containing 0.1 mole of a non-electrolyte solute below the vapor pressure of pure water.
per liter at 273 K. 4. The BP of CHCl3 was raised by 0.323 C0 when
0.37 g of naphthalene was dissolved in 35 g of
11. At 10 0C, the osmotic pressure of urea solution
CHCl3. Calculate the molecular weight of
was found to be 500 mm. The solution is naphthalene. For CHCl3, Kb = 3.9 K mol – 1 kg.
diluted and the temperature is raised to 25 0C 5. Pure benzene boils at 80 0C. The boiling point
where osmotic pressure was noticed to be of a solution containing 1 g of substance
105.3 Determine the extent of dilution. dissolved in 83.4 g of benzene is 80.175 0C. If
12. Osmotic pressure of a solution containing 2 g latent heat of vaporization of benzene is 90
of dissolved protein per 300 mL of solution is cal/g, calculate the molecular weight of
solute
 16

6. Calculate the FP of a 5 % glucose (by mass) in that expected for a nonelectrolyte

water.  Example
7. What are the molecular mass and molecular
formula of a non-electrolytic molecular  electrolyte   nonelectrolyte
compound with an empirical formula C4H2N if Tf
 3.72 for HCl, NH4 Cl
3.84 g of the compound in 500 g of benzene m
gives a freezing point depression of 0.307 C0? = 5.58 for CoCl3
8. What is the depression in FP of a solution of = 7.44 for K 3 Fe  CN6
non-electrolyte if the elevation in BP is 0.13 K.
 To describe this effect, defined the van’t Hoff
Kb = 0.52 and Kf = 1.86 K mol – 1 kg
9. Two elements A and B form compounds factor, i:
having molecular formula AB2 and AB4. When  measured value for the 
dissolved in 20 g benzene, 1 g of AB2 lowers  
 soln of electrolyte 
the freezing point by 2.3 C0 whereas 1.0 g of i 
 expected value for the 
AB4 lowers it by 1.3 C0. Kf for benzene is 5.1 K  
 soln of non-electrolyte 
mol – 1 kg. Calculate the atomic weights of A
Tf Tb P 
and B. i    
10. The osmotic pressure of blood is 7.65 atm at
 Tf 0  Tb  0  P 0  0
37 0C. How much glucose should be used per  electrolyte  iCRT Tf  imK f
liter for an intravenous injection that is to
have the same osmotic pressure as blood. Tb  imK b P  ixsolute Psolvent
0

11. At 25 0C, a solution containing 0.2 g of Tf  i  kf m    im  kf

polyisobutylene in 100 mL of benzene
developed a rise of 2.4 mm at osmotic mt
im  mt  i
equilibrium. Calculate the molecular weight m
of polyisobutylene if the density of solution is
0.88 g/mL. Van’t Hoff Factors at 0.05m Concentration in
12. A 1.1 g sample contains only glucose and Aqueous Solution
sucrose. When the sample is dissolved in
water to a total solution volume of 25 mL, the Solute Expected i Observed i
osmotic pressure of the solution is 3.78 atm NaCl 2 1.9
at 298 K. What is the % composition of MgSO4 2 1.3
glucose and sucrose in the sample? MgCl2 3 2.7
13. At 270C, a 5 % solution (mass/volume, that is, K2SO4 3 2.6
100 mL of the solution contains 5 g of solute) FeCl3 4 3.4
of cane sugar is isotonic with 8.77 g/L of urea
solution. Find the molar mass of urea if MW
of sugar is 342. Also report the osmotic For Electrolytes that Dissociate in Solution
pressure of solution if 100 mL each are mixed
1. If electrolyte is strong:
at 270C.
i  no. of ions per formula unit
 xy

COLLIGATIVE PROPERTIES OF ELECTROLYTES

2. If electrolyte is weak:
van’t Hoff
 Observed that certain solutions produce
greater effect on colligative properties than
 17

Ax By  xAy   yB x  1. Consider the following aqueous solutions:

n 0 0 i. 0.1 m Na3PO4 iii. 0.02 m KCl
ii. 0.2 m CaBr2 iv. 0.02 m HF
n xn yn
a. Assuming complete dissociation of the
n  n xn yn
soluble salts, which solution (s) would have
nt  n  n  xn  yn
the same BP as 0.04 m C6H12O6 in water?
n n  n  xn  yn
i t  b. Which solution would have the highest
n n
vapor pressure at 28 0C?
 1    x  y  1    x  y  1 
c. Which solution would have the lowest FP?
i 1 i 1
  where: v  x  y 2. From the following:
x  y 1 v 1 i. Pure water
Tf Tb
1 1  ii. 0.01 C12H22O11 (aq)
mk f 1
mkb CRT iii. 0.01 m NaCl (aq)
  
v 1 v 1 v 1 iv. 0.01 m CaCl2 (aq)
Which one will have the
For molecules undergoing association: a. Highest FP? d. Lowest BP?
 i<1 b. Lowest FP? e. Highest 
 Degree of Association,  c. Highest BP?
fraction of the total number of molecules
which associate together resulting in the Exercise
formation of bigger molecules 1. Calculate the mass of KCl which must be
mA  Am added to 1 kg of water so that the FP is
n 0 depressed by 2 K. Assume that KCl undergoes
n complete dissociation
 n 2. 1.2 % solution of NaCl is isotonic with 7.2 %
m
n solution of glucose (MW = 180). Calculate the
n  n degree of dissociation and van’t Hoff factor for
m
n NaCl solution.
nt  n  n  i
m 3. Average osmotic pressure of human blood is
n 7.4 atm at 27 0C. Calculate the total molarity
n  n 
i m    i 1 of various solutes. Assuming this oncentration
n 1 to be approximately equal to molality,
1
m calculate the FP of blood.
where: m = no. of molecules required to 4. A 0.10 M solution of HgCl2 in water freezes at
form one molecule – 0.186 0C whereas 0.10 M solution of
Example: if a solute has a tendency to dimerize Hg(NO3)2 freezes at – 0.465 0C. Find the value
(m = 2) and all its molecules undergo of the van’t Hoff factor I in these two cases.
association then What do these values tell us about the
i 1 structural states of these salts? Assume

1 M=m
1
2
Check Questions
 18

5. Ka of -chloroacetic acid is 0.00137. What is

the freezing point of 0.01 molal aqueous Problem Set
solution of the acid? 1. Assuming complete dissociation, calculate the
6. A sample of water is found to contain 5.85 % expected FP of a solution prepared by
by mass, NaCl and 9.5 % by mass, MgCl2. dissolving 6.00 g of Glauber’s salt Na2SO4 10
What is the boiling point of water sample H2O in 1 kg of water.
assuming 80 % ionization of NaCl and 50 % 2. A solution containing 2.47 g of ethyl benzoate
ionization of MgCl2? in 100 g of benzene had density 0.8 g/mL and
7. What is the FP of 0.2 molal solution of acetic vapor pressure 742.6 mm Hg at 80 0C. The
acid in benzene in which it dimerizes to the vapor pressure of pure benzene at same
extent of 60%. FP of benzene is 278.4 K and temperature is 751.6 mm Hg. What is the
its molar heat of fusion is 10.042 kJ/mol osmotic pressure of the solution?
8. An aqueous solution containing 0.25 mol Q, a 3. 2 L of ammonia at 27 0C and 0.2 bar pressure
strong electrolyte, in 500 g water freezes at – is required to completely neutralize 150 mL of
2.79 0C. H2SO4 solution. What is the molarity of this
a. What is the van’t Hoff factor for Q? acid? Also calculate the osmotic pressure of
b. What is the formula of Q if it is 38.68 the sulfuric acid at 27 0C.
% 4. An acid with MW = 300 is dissolved in water to
chlorine by mass and there are twice as make 0.1 M solution having density 1.01 g/mL.
many anions as cations in one formula If the acid is 4.5 % ionized, calculate the FP of
unit of Q? the solution
9. 1.22 g of benzoic acid is dissolved in 5. 0.003 kg of acetic acid is added to 500 mL of
i. 100 g of acetone (Kb = 1.7) water. If 23 % of acetic acid is dissociated,
ii. 100 g of benzene (Kb = 2.6) what will be the depression in FP? Density of
The elevations in BP is 0.17 C0 and 0.13 C0 water is 0.997 g/mL.
respectively. What are the molar masses of 6. The FP of a solution containing 0.2 g of acetic
benzoic acids in 2 solvents? acid in 20 g of benzene is lowered by 0.45 C0.
a. What do you deduce of it in terms of Calculate the degree of association of acetic
structure of benzoic acid? acid in benzene (Kf for benzene of 5.12 K
b. What do you deduce of it in terms of Kg/mol.
structure of benzoic acid? 7. A storage battery contains a solution of
10. What is the ratio by weight of NaF and NaI sulfuric acid 38 % by weight. At this
which when dissolved in water produces the concentration, the van’t Hoff factor is 2.50. At
same osmotic effects as 0.1 molar solution of what temperature will the battery contents
urea in water? The weight of residue freeze? Kf = 1.86 K/molal
obtained on evaporation of the salt solution 8. The freezing point depression of a 0.091 m
is 0.48 gram per 100 mL of solution solution of CsCl is 0.320 C0. That of a 0.091 m
evaporated. Assume complete solution of CaCl2 is 0.440 . In which solution
dissociation of the salts. does ion association appear to be greater?
Explain.
9. An aqueous solution is 1 % NaCl by mass and
has a density of 1.071 g/mL at 25 0C. The
 19

observed osmotic pressure of this solution is

7.83 atm at 25 0C.
a. What fraction of the moles of NaCl exist
as ion-pairs?
b. Calculate the FP of this solution.
10. A certain mass of a substance, when
dissolved in 100 g benzene, lowers the
freezing point by 1.28 C0. The same mass of
solute dissolved in 100 g of water lowers the
freezing point by 1.4 C0. If the substance has
normal molecular weight in benzene and is
completely ionized in water, into how many
ions does it dissociate in water? Kf for
benzene = 5.12 K /mol/kg.
11. The degree of dissociation of Ca(NO3)2 in a
dilute aqueous solution containing 7 g of salt
per 100 g of water at 100 0C is 70 %. Calculate
the vapor pressure of the solution.
12. Calculate the molality of an NaCl solution
which exhibits elevation in BP numerically
equal to the depression in FP of 0.1 molal
aqueous solution of aluminum sulfate.
Assume complete dissociation of the salts.
13. Calculate the FP of an aqueous solution of a
non-electrolyte having osmotic pressure of 2.0
bar at 300 K. Assume molarity to be
approximately equal to molality.