Solution of non electrolyte full

What is a solution?
• A solution is a homogeneous mixture
• A solute is dissolved in a solvent.
• Solute is the substance being dissolved
• Solvent is the medium in which the solute dissolves
• An aqueous solution has water as solvent
2

SATURATED – contains the
max. no. of solute that
dissolves at that temperature
SUPERSATURATED –
Contains more than is
possible and are unstable
Unsaturated solution Saturated solution
Solution
Unsaturated Saturated
Supersaturated
3

Super saturated solution
Solvent holds more solute than is normally possible
at that temperature.
These solutions are unstable; crystallization can
often be stimulated by adding a “seed crystal” or
scratching the side of the flask.
4

How Does a Solution Form?
1. Solvent molecules attracted to surface ions.
2. Each ion is surrounded by solvent molecules.
3. Enthalpy (DH) changes with each interaction broken or
formed.
5

It is a 3 step process
1. Separation of Solute
• Must overcome IMF or ion-ion attractions in solute
• Requires energy, ENDOTHERMIC ( + DH)
2. Separation of Solvent
• Must overcome IMF of solvent particles
• Requires energy, ENDOTHERMIC (+ DH)
3. Interaction of Solute & Solvent
• Attractive bonds form between solute particles and solvent
particles
• “Solvation” or “Hydration” (where water = solvent)
• Releases energy, EXOTHERMIC (- DH)
6

WAYS OF EXPRESSING
CONCENTRATIONS OF SOLUTIONS
8

Mass Percentage
100)/%( x
solutionofmass
soluteofmass
ww 
100)/%( x
solutionofvolume
soluteofmass
vw 
100)/%( x
solutionofvolume
soluteofvolume
vv 
𝑼𝒏𝒊𝒕 % 𝒐𝒇 𝑨 =
𝑼𝒏𝒊𝒕 𝒐𝒇 𝑨 𝒊𝒏 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝑼𝒏𝒊𝒕 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝑿 𝟏𝟎𝟎
9

What is the per cent by weight of NaCl if 1.75 g
of NaCl is dissolved in 5.85 g of water
• Wt. of solute (NaCl) = 1.75 g
• Wt. of solvent (H2O) = 5.85 g
• ∴ Wt. of solution = 1.75 + 5.85 = 7.60 g
• Hence concentration of NaCl % by weight
• 1.75/ 7.60 X 100 = 23.0
10

Parts per Million (ppm) and
Parts per Billion (ppb)
10
9
solutionofmassTotal
solutioninAof
XMass
ppb 
10
6
solutionofmassTotal
solutioninAof
XMass
ppm 
11

Mole Fraction (X)
moles of A
total moles in solution
XA =
12

• If n represents moles of solute and N number of moles of
solvent,
𝑋𝑠𝑜𝑙𝑢𝑡𝑒 =
𝑛
𝑛 + 𝑁
• Notice that mole fraction of solvent would be
𝑋𝑠𝑜𝑙𝑣𝑒𝑛𝑡 =
𝑁
𝑛 + 𝑁
• Mole fraction is unitless and 𝑋𝑠𝑜𝑙𝑢𝑡𝑒 + 𝑋𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = 1
13

Calculate the mole fraction of HCl in a solution of hydrochloric
acid in water, containing 36 per cent HCl by weight
14

Molarity (M)
• Because volume is temperature dependent,
molarity can change with temperature.
• Unit = mol/ litre
mol of solute
L of solution
M =
15

What is the molarity of a solution prepared by
dissolving 75.5 g of pure KOH in 540 ml of solution.
16

What weight of HCl is present in 155 ml of a 0.540 M
solution
17

Molality (m)
• Because neither moles nor mass change with
temperature, molality (unlike molarity) is NOT
temperature dependent.
• Unit = mol/ kg
mol of solute
kg of solvent
m =
18

What is the molality of a solution prepared by dissolving 5.0 g
of toluene (C7H8) in 225 g of benzene (C6H6) ?
19

Normality (N)
• The normality of a solution is the gram equivalent weight
of a solute per litre of solution.
• Normality is the only concentration unit that is reaction
dependent.
solutionofLitre
soluteofweightequivalentGram
Normality
20

Equivalent weight (g/Eq)
• It is the mass of one equivalent, that is the mass of a
given substance which will:
• Supply or react with one mole of hydrogen cations H+ in
an acid–base reaction; or
• Supply or react with one mole of electrons e− in a redox
reaction.
• It is that weight of any atom/molecule which displaces
1.008 g of H, 19 g of F or 8 g of O.
/moleequivalent
weightmolecular
weightEquivalent
21

5 g of NaCl is dissolved in 1000 g of water. If the density of the
resulting solution is 0.997 g/ml, calculate the molality,
molarity, normality and mole fraction of the solute,
assuming volume of the solution is equal to that of solvent
22

Avogadro Number
• Avogadro hypothesized that there was a specific number
that would represent the number of atoms or molecules in
a mole of that atom or molecule.
• The weight of that unit known as a mole would be
equivalent to the molecular weight of the atom or
molecule in grams. (Mole = Molecular weight in grams)
26

• According to this theory, one mole of carbon-12 would have a
mass of 12 grams because carbon-12 has an atomic weight of
12.
• One mole of hydrogen would weigh one gram
• It would contain the same number of atoms as one mole of
carbon.
• The magical number was, in fact, discovered to be 6.023E23
(6.023 X 1023)
27

SOLUTIONS OF LIQUID IN
LIQUID
Ideal and Real solutions
Raoult’s Law
Deviation from Raoult’s law
28

Solutions of liquid in liquid
Liquid pairs
Completely
miscible
Partially
miscible
Completely
immiscible
Real
solutions
Ideal
solutions
29

Ideal solutions
Cohesion Adhesion
30

• No heat is evolved or absorbed during
mixing
• Final volume of sol.= sum of both.
• The properties of solution such as vapour
pressure, surface tension, viscosity etc are
the average of the two pure liquids.
31

Escaping tendency
Quantitative measure:
 Pure substance: Molar free energy
 Constituent of a solution: Partial molar free energy or
chemical potential
32

Ideal mixtures and intermolecular
forces
• In a pure liquid, some of the
more energetic molecules have
enough energy to overcome the
intermolecular attractions and
escape from the surface to form a
vapour.
• The smaller the intermolecular
forces, the more molecules will
be able to escape at any
particular temperature
Liquid A
33

• If you have a second liquid, the same thing is true.
• At any particular temperature a certain escaping tendency
Liquid B
34

• In an ideal mixture of these two liquids.
• There will be equal evaporation and hence equal vapour
pressure
1:1
Solution of A + B
35

Vapour pressure
• Vapour pressure or equilibrium vapour pressure is the pressure of a
vapour in thermodynamic equilibrium with its condensed phases in a
closed system.
3:2
36

Raoult’s Law (1887)
• Partial vapour pressure of each volatile constituent is equal to the
vapour pressure of the pure constituent multiplied by its mole
fraction in the solution. Thus, for two constituents A and B,
PA = PA° XA
PB = PB° XB
PA and PB – partial vapour pressure
XA and XB – mole fraction concentration
PA° and PB° - vapour pressure of pure components
• The total vapour pressure of the mixture is equal to the sum of
the individual partial pressures.
Total Vapour Pressure = PA + PB
37

• E.g. if vapour pressure of ethylene chloride in the pure
state is 236 mm Hg at 50°C, then in a solution consisting
of a mole fraction of 0.4 ethylene chloride and 0.6
benzene, the partial vapour pressure of ethylene chloride
is 40% of 236 mm.
PA = PA° XA
Pec = Pec° Xec
Pec = 236 X 0.4
Pec = 94.4 mm
38

The presence of a non-volatile solute means that fewer solvent
particles are at the solution’s surface, so less solvent evaporates!
Escaping tendency decreases
39

• Thus, in an ideal solution, when liquid A is
mixed with liquid B in a manner depending on
the mole fractions of A and B present in the
final solution.
• This will diminish the escaping tendency of
each constituent, leading to a reduction in the
rate of escape of the molecules of A and B from
the surface of liquid.
• The total pressure is the sum of the partial
pressures of all the constituents.
P = PA + PB
40

Real Solutions
Cohesion
Adhesion
Do not adhere to
Raoult’s law
Cohesion
Adhesion
OR
42

Deviation from Raoult’s Law
Positive Deviation Negative deviation
43

Negative deviation
Cohesion
Adhesion
Vapour pressure of
solution less than expected
These are cases where
the molecules break
away from the mixture
LESS easily than they do
from the pure liquids.
New STRONGER forces
must exist in the mixture
than in the original
liquids.
A↔B 〉〉 A↔A, or B↔B
44

Chloroform (A)
Acetone (B)
Solution of acetone in
chloroform
45

• Dilution of chloroform (A) by the addition of acetone (B).
• Addition of B to A tends to reduce the vapour pressure of A to a
greater extent than can be accounted for by the simple dilution.
• Chloroform and acetone manifest such attraction for one
another through the formation of a hydrogen bond further
reducing the escaping tendency of each constituent.
47

This pair forms weak compound, Cl3C-H…O=C(CH3)2
which can be isolated and identified.
HC
Cl
Cl
Cl
O C
CH3
CH3
Reaction between dipolar molecules, or between a dipolar
and a non polar molecule, may also lead to negative
deviations.
48

Positive deviation
Vapour pressure of
solution greater than
expected
These are cases where the
molecules break away
from the mixture
MORE easily than they do
from the pure liquids. New
WEAKER forces
must exist in the mixture
than in the original liquids.
A↔B << A↔A, or B↔B
Cohesion
Adhesion
49

Ethyl alcohol (A)
Chloroform (B)
Solution of chloroform in
ethyl alcohol
Benzene + Ethyl
alcohol
Carbon
disulphide +
Acetone
Other
examples
50

• Raoult’s law describes the behavior of either of the
component of a real liquid pair only when that substance
is present in high concentration and thus is considered to
be the solvent.
• In such a situation Raoult’s law can be expressed as
Psolvent = P°solvent Xsolvent
• It is valid only for the solvent of a nonideal solution that
is sufficiently dilute with respect to the solute. It cannot
hold for the component in low concentration, that is, the
solute in a dilute solution.
52

Limitations of Raoult’s
law
Real
solutions
In real solutions the concentration of solute is high
and the intermolecular forces between solute-solute
and solute-solvent are high. This causes deviations.
Volatile
solute
It is applicable to only non-volatile solute. As volatile
solutes contribute to the vapour pressure which may
cause deviation.
Solutes
which
associate or
dissociate
If it associates it leads to reduction in lowering of
vapour pressure and if it dissociates then vapour
pressure lowering would be increased.
53

Lowering of vapour
pressure
• According to Raoult’s law, the vapour pressure of a
solvent over a dilute solution is equal to the vapour
pressure of pure solvent multiplied by its mole fraction
𝑝 = 𝑝1
𝑜 𝑋1---------------------- 1
• Because the solute is non-volatile, the vapour pressure of
the solvent is identical to the total vapour pressure of the
solution
54

• It is more convenient to express vapour pressure of
the solution in terms of the concentration of the
solute rather than the mole fraction of the solvent,
the conversion maybe achieved as follows:
𝑋1 + 𝑋2 = 1
𝑋1 = 1 − 𝑋2-------------------2
• Where, X1 = mole fraction of the solvent
X2 = mole fraction of the solute
• Substituting 2 in 1, we get
p = 𝑝1
𝑜
(1 − 𝑋2)
𝑝0
1 − p = 𝑝1
0 𝑋2
55

𝑝1
𝑜 − 𝑝
𝑝1
𝑜
=
∆𝑝
𝑝 𝑜
1
= 𝑋2 =
𝑛2
𝑛1 + 𝑛2
∆𝑝
𝑝 𝑜
1
=
𝑤2
𝑀2
𝑤2
𝑀2
+
𝑤1
𝑀1
In the above eq.
ΔP = P1° – P is the lowering of the
vapour pressure; and
ΔP / P1° is the relative vapour
pressure lowering.
56

• If the concentration of the solute is very less, its
number of moles in the denominator can be
ignored, thus the equation would become:
∆𝑝
𝑝 𝑜
1
=
𝑤2
𝑀2
𝑤1
𝑀1
∆𝑝
𝑝 𝑜
1
=
𝑤2 𝑀1
𝑤1 𝑀2
Vapour pressure lowering can be used to calculate
molecular weight of a compound
57

AEROSOLS
• Uses Raoult’s law.
• It has a drug and a propellant.
• Common propellants used:
• Trichloromonofluoromethane (propellant 11)
• Dichlorodifluoromethane (propellant 12)
58

Henry’s Law
• The effect of partial pressure on solubility of gases
• At pressure of few atmosphere or less, solubility of gas
solute follows Henry Law which states that the amount of
solute gas dissolved in solution is directly proportional to
the amount of pressure above the solution
c = k P
c = solubility of the gas (M)
k = Henry’s Law Constant
P = partial pressure of gas
59

Henry’s Law & Soft Drinks
• Soft drinks contain “carbonated water” –
water with dissolved carbon dioxide gas.
• The drinks are bottled with a CO2 pressure
greater than 1 atm.
• When the bottle is opened, the pressure of
CO2 decreases and the solubility of CO2
also decreases, according to Henry’s Law.
• Therefore, bubbles of CO2 escape from
solution.
61

Henry’s law applies to the
SOLUTE and
Raoult’s law applies to the
SOLVENT
in dilute solutions of real liquid
pairs.
62

COLLIGATIVE PROPERTIES OF
SOLUTIONS
63

Properties of solutions
• Solutions have properties different from both solute and
solvent
• 4 types of properties
64
Additive Constitutive Colligative
Additive &
Constitutive

Additive properties
• Additive Properties: Additive properties are those
properties which is the sum of the corresponding
properties of the atoms constituting the molecule.
• These properties only depend on the types of the atom
and their numbers
• e.g., mass is a additive property, similarly molar volume
is also a good example of additive properties.
65

Constitutive properties
• Constitutive property of a molecule is the property which
depends upon the constitution of the molecule
• i.e., upon the arrangements of atoms within the molecule
e.g.,
• Optical activity.
66

Additive and constitutive
properties
• The physical property which depend upon the number of
atom in a molecule as well as their constitution
• e.g., atomic volume, parachor etc.
67

Colligative properties
• Colligative properties are those properties, which depends
upon the number of molecules present in a substance
• e.g., vapour pressure of gas, elevation in boiling point,
depression of freezing point, osmotic pressure of the
solution, etc.
68

Lowering of vapour
pressure
69

Lowering of vapour pressure
• Pressure is measured with a manometer
• When a non volatile solute is combined with a
volatile solvent, the vapour above the solution is
provided by the solvent only.
• Solute reduces the escaping tendency of solvent.
• Vapour pressure of a solution containing a non
volatile solute is lowered proportional to the relative
number of the solute molecules.
• Therefore the vapour pressure of the solvent, P1 is
identical to the total pressure of the solution, P.
70

The presence of a non-volatile solute means that fewer solvent
particles are at the solution’s surface, so less solvent evaporates!
Escaping tendency decreases
71

• It is more convenient to express the vapour pressure
of the solution in terms of concentration of solute
rather than the mole fraction of solvent
• The sum of the mole fractions of the constituents in
a solution is unity:
X1 + X2 = 1
X1 = 1 – X2
Where,
• X1 is the mole fraction of the solvent; and
• X2 is the mole fraction of the solute.
72

Raoult’s eq.
73
In the above eq.
ΔP = P1° – P is the lowering of the
vapour pressure; and
ΔP / P1° is the relative vapour
pressure lowering.

Determination of
vapour pressure of
solutions
1. Manometer - Isopiestic method
2. Hill and Baldes Apparatus
3. Wescor Method
74

Methods for determination
of vapour pressure lowering
Methods
Static
Barometric
method
Manometer
– isopiestic
method
Hill and
Baldes
method
Wescor
method
Dynamic
Ostwald
Walker
75

Barometric Method
• Raoult measured the individual
vapour pressure of a liquid and
then the solution by this method.
• He introduced the liquid or the
solution into Toricellian vacuum of
a barometer tube and measured the
depression of the mercury level.
• This method is neither practicable
nor accurate as the lowering of
vapour pressure is too small.
76

Manometer
• Vapour pressure lowering is obtained by
subtracting the vapour pressure of the solution
from the vapour pressure of the pure solvent.
77Vapour pressure
of solution
Vapour
pressure
lowering
Vapour pressure
of solvent

78Apparatus for the isopiestic method
The vapour pressure of KCl solution of various
concentrations have been determined accurately
and thus the vapour pressure of the test solution
that is isopiestic is thus readily obtained.
Isopiestic method is used for precise determination of vapour pressures.

It consists of combination
of wires of different alloys
formed onto two loops and
connected to a
galvanometer.
Used for determining the
relative vapour pressure of
small amounts of liquids.
This thermoelectric method
measure the change in
potential with respect to
change in vapour pressure.
The solution of known
vapour pressure and an
unknown evapourate in a
chamber maintained at
constant humidity.
Vapour pressure lowering
of solution is then obtained
from a standard curve of
vapour pressure versus
galvanometer readings of
potential.
This method is used to
study the colligative
properties of ophthalmic
solutions.
80

Wescor vapour pressure osmometer.
• It is the fastest and easiest method of determining
osmolality.
• Therefore it is the method of choice for most of the fluids
in biology and medicine in which water is the solvent.
• The test solution is absorbed onto a filter paper disk
which is usually 2 to 10 µL.
• The disk is placed in a sealed chamber near the
thermocouple, which is cooled below the dew point of the
solution.
• Thermocouple is then equilibrated to the dew point of the
solution whereupon its potential is recorded.
• The potential determined is proportional to the vapour
pressure lowering.
• Reference standard solutions are used to calibrate the
potential readings against known vapour pressures at the
ambient temperature.
81

• This instrument has been applied to quantitating sodium in
isotonic solutions and studying the colligative properties of
parenteral solutions.
• This instrument is also called as vapour pressure
differentiometers as it does not involve membrane diffusion
operation.
83

Ostwald and Walker’s Dynamic
Method (Gas Saturation Method)
• In this method the relative lowering of vapour pressure can be determined
straightway.
• The measurement of the individual vapour pressures of a solution and
solvent is thus eliminated.
• Procedure. The apparatus used by Ostwald and Walker is shown in Fig. It
consists of two sets of bulbs :
(a) Set A containing the solution
(b) Set B containing the solvent
• Each set is weighed separately. A slow stream of dry air is then drawn by
suction pump through the two sets of bulbs.
• At the end of the operation, these sets are reweighed.
• From the loss of weight in each of the two sets, the lowering of vapour
pressure is calculated.
• The temperature of the air, the solution and the solvent must be kept
constant throughout.
84

• Knowing the loss of mass in set B (w2) and the total loss of
mass in the two sets (w1 + w2), we can find the relative
lowering of vapour pressure from equation (4).
• If water is the solvent used, a set of calcium chloride tubes
(or a set of bulbs containing conc. H2SO4) is attached to the
end of the apparatus to catch the escaping water vapour.
• Thus the gain in mass of the CaCl2-tubes is equal to (w1 +
w2), the total loss of mass in sets A and B.
87

VAPOUR PRESSURE OF LIQUID-
LIQUID SOLUTIONS
88

• The study of the vapour pressures of mixtures of
completely miscible liquids has proved of great help in
the separation of the liquids by fractional distillation.
• The vapour pressures of two liquids with varying
composition have been determined at constant
temperature.
• By plotting the vapour pressure against composition it
has been revealed that, in general, mixtures of the
miscible liquids are of three types.
89

Miscible liquids
1. First Type of Mixtures of Miscible
Liquids (Maximum boiling point
azeotropic solutions)
2. Second Type of Mixtures of
Miscible Liquids (Minimum boiling
point azeotropic solutions)
3. Third Type of Mixtures of Miscible
Liquids
90

Azeotrope / Azeotropic Mixture
• Very large deviations from ideality lead to a special class of
mixtures known as azeotropes, azeotropic mixtures, or
constant-boiling mixtures.
• Azeotrope is a special class of liquid mixture that boils at a
constant temperature at a certain composition.
• At this condition, it behaves as if it was one component with
one constant boiling point.
92

First Type of Mixtures of Miscible
Liquids (Maximum boiling point
• For this type of solutions the vapour pressure curve
exhibits a minimum.
• If we take a mixture which has an excess of X (more
volatile component), we are somewhere at C on the
curve.
• When this is distilled the vapour will contain excess of X
and thus the remaining mixture will get richer in Y.
• Finally we reach the point D where vapour pressure is
minimum and thus boiling point is maximum.
• Here the mixture will distil unchanged in composition.
93

• It is obvious that complete separation of this type of
solutions into components is impossible.
• At best it can be resolved into one pure component and
the constant boiling mixture.
• Solutions of this type which distil unchanged at a
constant temperature and show a maximum boiling point
are called maximum boiling point azeotropic solutions.
96

Maximum boiling
azeotropes
 It occurs when the negative deviations are very large
 The total pressure curve in this case passes through
a minimum, giving rise to a maximum in
the temperature (i.e. boiling point)
 EXAMPLES
 Hydrochloric acid - Water (11.1 mole% HCl, 110 oC, 1
atm)
 Acetone - Chloroform (65.5 mole% chloroform, 64.5 oC,
1 atm)
 Nitric acid – Water (68 mole% HNO3 120 oC, 1 atm)
97

Second Type of Mixtures of Miscible
Liquids (Minimum boiling point
• Ethanol and water mixtures offer a good example of this
type.
• Ethanol-water mixture containing 95.6 per cent ethanol
boils at the minimum temperature 78.13°.
• Thus it is very difficult to obtain pure absolute alcohol by
distillation.
• This difficulty has, however, been overcome by adding
benzene which form a low boiling mixture with water and
on distillation it comes over leaving pure ethanol behind.
98

• OTHER EXAMPLES
• Ethanol-Water system which at 1 atm occurs at 89.4
mole percent ethanol and 78.2 oC.
• Carbon disulfide – Acetone 61.0 mole% CS2, 39.25 oC,
1 atm
• Benzene - Water (29.6 mole% H2O, 69.25 oC, 1 atm)
101

Third Type of Mixtures of
Miscible Liquids
• In this case the vapour pressures of mixtures always lie between
the vapour pressures of pure components and thus the vapour-
pressure composition curve is a straight line.
• Suppose we have a mixture containing excess of Y which is
represented by point G on the curve.
• On distillation X component being more volatile will be obtained
in greater proportion in the distillate and we gradually travel along
the curve AB.
• The latter fractions will, of course, be poorer in X and richer in Y
till we reach the 100 per cent Y-axis, when all the X will have
passed over.
102

• Only in this type of solutions we can completely separate
the components by fractional distillation.
• Thus methyl alcohol-water mixtures can be resolved into
pure components by distillation.
• Liquid mixtures which distil with a change in
composition are called zeotropic mixtures
103

Distillation
 Distillation is a widely used method for separating
mixtures based on differences in the conditions required
to change the phase of components of the mixture.
 To separate a mixture of liquids, the liquid can be heated
to force components, which have different boiling points,
into the gas phase.
 The gas is then condensed back into liquid form and
collected.
 Vapourization of a liquid and subsequent condensation
of the resultant gas back to liquid form – Distillation
104

Boiling point
 The boiling point of an element or a substance is the
temperature at which the vapour pressure of the liquid equals
the environmental pressure surrounding the liquid (760 mm
Hg)
Atmospheric
pressure
Vapour
pressure
Boiling
point
107

Volatile
Vapour
pressure
Boiling
point
108

USING BP TO SEPARATE A
MIXTURE OF 2 LIQUIDS
110

Miscible liquids – fractional
distillation (Theory)
• To understand the process of fractional distillation we must
have an idea of the composition of the vapour phase and that
of the liquid mixtures at different boiling temperatures.
• Thus for this purpose it is not the vapour-pressure composition
curve but rather the temperature-composition curve that is
important.
• If we plot the boiling point of liquid mixture against its
composition and the composition of the vapour in contact with
it, we get two separate curves for each type of solutions.
• The curves obtained for the third type are shown
111

The curves AEB and ADB
are the temperature
composition curves for the
vapour and liquid
respectively.
At any boiling temperature
C the composition of liquid
mixture is represented by J
and that of the vapour in
equilibrium by K.
Obviously, the more volatile
component Y is present in
greater proportion in the
vapour than the liquid
mixture.
Thus the condensed vapour
or the distillate will be
richer in Y.
If the distillate so obtained
be now subjected to
distillation, it will boil at F
and the fresh distillate will
have the composition L
corresponding to I.
Thus the proportion of Y in
the second distillate is
greater than in the first one.
In this way by repeating the
process of fractional
distillation it is obvious that
we can get almost pure Y.
113

• In first type of solutions if we have a boiling mixture
represented by Y its vapour will be poor in Y than the liquid
mixture and the boiling point would gradually rise till we reach
the maximum point C where the composition of liquid and
vapour is the same.
• Here the distillation proceeds without change of composition.
• Similarly in the second type, if we have a boiling mixture
represented by the point X', the amount of Y in vapour is higher
and gradually the boiling point falls to the minimum C' where
the vapour and the liquid mixtures have the same composition.
• At this temperature the mixture boils without any change in
composition.
• Thus it is proved that the second and first type of solutions
are not capable of being separated by fractional
distillation. 115

Apparatus for fractional
distillation
• The efficiency of the process of fractional distillation is considerably
enhanced by the use of the so-called Fractionating Columns.
• These are of different designs.
• An effective and simple fractionating column usually employed for
laboratory use consists of a long glass tube packed with glass beads or
specially made porcelain rings.
• The glass tube blown into bulbs at intervals may also constitute a
fractionating column.
• For industrial purposes a fractionating tower is employed.
• A fractionating tower is divided into several compartments by means of
tray that are set one above the other.
• There is a hole in the centre of every tray which is covered by bubble cap.
• Each tray has an overflow pipe that joins it with the tray below by
allowing the condensed liquid to flow down
116

The fractionating column or
tower is fitted in the neck of the
distillation flask or the still so
that the vapours of the liquid
being heated pass up through it.
The temperature falls in the
column as vapours pass from
bottom to the top.
The hot vapours that enter the
column get condensed first in
the lowest part of it.
As heating is continued more
vapours ascend the column and
boil the liquid already
condensed, giving a vapour
which condenses higher up in
the column.
This liquid is heated in turn by
more vapours ascending the
column.
Thus the liquid condensed in
the lowest part is distilled on to
the upper part.
In this manner a sort of
distillation and condensation
goes on along the height of the
column which results in the
increase of the proportion of the
volatile component in the
outgoing vapours.
At every point in the column
there exists an equilibrium
between liquid and vapour.
This is established quickly by
and upward flow of vapours
and the downward flow of
liquid, a large surface area and
a slow rate of distillation.
118

• A simple
distillation of
a mixture of
methanol and
water and the
liquid vapour
equilibrium
states are
depicted
119

• It is clear that the liquid-vapour equilibria change
regularly in moving up the column.
• We may withdraw mixtures of varied compositions from
different points on the column.
• This is done in the fractional distillation of crude oil in a
refinery where different products of industrial use are
conveniently separated.
120

Immiscible liquids
• When 2 immiscible liquids are heated while being agitated,
each constituent independently exerts its own vapour pressure
as a function of temperature as the other liquid does not exist
• Boiling begins when the sum of the partial pressures of the two
liquids just exceeds the atmospheric pressure
• This principle is used in steam distillation
121

Steam Distillation
• Organic substances insoluble in
water can purified
• Water insoluble substances can
separated at temperature below
their degradation temperature
Water
1oo ◦C
Bromo-benzene
156.2 ◦C
Mixture
95 ◦C
Useful for separating volatile oils from plant tissue without
decomposing the oils
122

Steam distillation
• Distillation carried in a current of steam is called steam
distillation.
• This technique is widely used for purification of organic
liquids which are steam volatile and immiscible with water
(e.g., aniline).
• The impure organic liquid admixed with water containing non-
volatile impurities is heated and steam passed into it.
• The vapour of the organic liquid and steam rising from the
boiling mixture pass into the condenser.
• The distillate collected in the receiver consists of two layers,
one of the pure organic liquid and the other of water.
• The pure liquid layer is removed by means of a separator
funnel and further purified.
124

Theory of steam
distillation
• The vapour pressure of a liquid rises with increase of
temperature.
• When the vapour pressure equals the atmospheric pressure, the
temperature recorded is the boiling point of the given liquid.
• In case of a mixture of two immiscible liquids, each
component exerts its own vapour pressure as if it were alone.
• The total vapour pressure over the mixture (P) is equal to the
sum of the individual vapour pressures (p1, p2) at that
temperature.
• P = p1 + p2
125

• Hence the mixture will boil at a temperature when the
combined vapour pressure P, equals the atmospheric
pressure.
• Since P > p1 or p2, the boiling point of the mixture of
two liquids will be lower than either of the pure
components.
126

• In steam distillation the organic liquid is mixed with water
(bp 100°C).
• Therefore the organic liquid will boil at a temperature lower
than 100°C.
• For example, phenylamine (aniline) boils at 184°C but the
steam distillation temperature of aniline is 98°C.
• Steam distillation is particularly used for the purification
of an organic liquid (such as phenylamine) which
decomposes at the boiling point and ordinary distillation
is not possible.
127

Elevation of boiling point
128

• Normal boiling point is the temperature at
which the vapour pressure of the liquid
becomes equal to an external pressure of 760
mm Hg.
• A solution will boil at higher temperature than
the pure solvent.
• The more the solute will be dissolved the
greater will be the boiling point elevation.
• The boiling point of a solution of a non volatile
solute is higher than that of the pure solvent
because the solute lowers the vapour pressure
of the solvent.
129

Greater solute
Greater elevation
Colligative
property
130

Relationship between Elevation of
Boiling Point and Lowering of Vapour-
pressure
• When a liquid is heated, its vapour pressure rises and when it
equals the atmospheric pressure, the liquid boils.
• The addition of a non volatile solute lowers the vapour
pressure and consequently elevates the boiling point as the
solution has to be heated to a higher temperature to make its
vapour pressure become equal to atmospheric pressure.
• If Tb is the boiling point of the solvent and T is the boiling
point of the solution, the difference in the boiling points (ΔT)
is called the elevation of boiling point.
• T – Tb = ΔT
131

• Consider the vapour pressure curves of
the pure solvent, and solutions (1) and
(2) with different concentrations of
solute
• For dilute solutions, the curves BD and
CE are parallel and straight lines
approximately.
• Therefore for similar triangles ACE
and ABD, we have
• where p – p1 and p – p2 are lowering
of vapour pressure for solution 1 and
solution 2 respectively.
• Hence the elevation of boiling point
is directly proportional to the
lowering of vapour pressure
• ΔT ∝ p – ps/p 132
Ostwald-Walker method of
measuring the relative lowering of
vapour pressure

Determination of Molecular Mass
from Elevation of Boiling Point 133

• where Kb is a constant called Boiling point constant or
Ebulioscopic constant of molal elevation constant.
• If w/m = 1, W = 1, Kb = ΔT.
• Thus, Molal elevation constant may be defined as the
boiling-point elevation produced when 1 mole of solute is
dissolved in one kg (1000 g) of the solvent.
• If the mass of the solvent (W) is given in grams, it has to be
converted into kilograms.
• Thus the expression (5) assumes the form
• where ΔT = elevation of boiling point;
• Kb = molal elevation constant;
• w = mass of solute in grams;
• m = mol mass of solute; and
• W = mass of solvent in grams.
• The units of Kb are °C kg/ mol
134

• Kb has a characteristic value for each solvent.
• It may considered as the boiling point elevation for an
ideal 1m solution.
• Kb is the ratio of the boiling point elevation to the molal
concentration in an extremely dilute solution in which
the system is approximately ideal.
135

Calculation of Kb using
thermodynamics 136

• After applying Clapeyron equation it is written as
137
Where,
• Vv and V1 are the molar volume of the gas and
the molar volume of the liquid,
• Tb is the boiling point of the solvent, and
• Δ Hv is the molar heat of vapourization.
Vv the volume of 1 mole of gas is replaced by RTb/P°
V1 is negligible compared to Vv the equation becomes
OR

but Δp/ P1° = X2 and this equation can be written as
138
where R = gas constant; Tb = boiling point of solvent; Hv = molar latent heat of
vaporization
Thus for water R = 8.134 J/mol; T = 373 K, Hv = 2260 J/g

Measurement of
elevation in boiling point
• Landsberger-Walker Method
• Cottrell’s Method
139

Landsberger-Walker
Method
• This method was introduced by Landsberger and modified by
Walker.
• Apparatus. The apparatus used in this method is shown and
consists of :
• (i) An inner tube with a hole in its side and graduated in ml;
• (ii) A boiling flask which sends solvent vapour in to the graduated
tube through a ‘rosehead’ (a bulb with several holes)’
• (iii) An outer tube which receives hot solvent vapour issuing from
the side-hole of the inner tube;
• (iv) A thermometer reading to 0.01 K, dipping in solvent or
solution in the inner tube.
140

141
• Pure solvent is placed in the graduated
tube and vapour of the same solvent
boiling in a separate flask is passed into it.
• The vapour causes the solvent in the tube
to boil by its latent heat of condensation.
• When the solvent starts boiling and
temperature becomes constant, its boiling
point is recorded.
• Now the supply of vapour is temporarily cut
off and a weighed pellet of the solute is
dropped into the solvent in the inner tube.
• The solvent vapour is again passed
through until the boiling point of the solution
is reached and this is recorded.
• The solvent vapour is then cut off,
thermometer and rosehead raised out of
the solution, and the volume of the solution
read.

• From a difference in the boiling points of solvent and
solution, we can find the molecular weight of the solute
by using the expression
• where w = weight of solute taken, W = weight of solvent
which is given by the volume of solvent (or solution)
measured in ml multiplied by the density of the solvent at
its boiling point.
143

Disadvantage
• Superheating:- The heating of the vapours of a solvent
which increases its temperature from its boiling point is
called superheating
• Complicated
144

Cottrell’s Method
• A method better than Landsberger-Walker method was
devised by Cottrell (1910).
• Apparatus. It consists of :
• (i) a graduated boiling tube containing solvent or solution;
• (ii) a reflux condenser which returns the vapourised solvent
to the boiling tube;
• (iii) a thermometer reading to 0.01 K, enclosed in a glass
hood;
• (iv) A small inverted funnel with a narrow stem which
branches into three jets projecting at the thermometer bulb.
145

Beckmann thermometer
• It is differential thermometer.
• It is designed to measure small changes in
temperature and not the temperature itself.
• It has a large bulb at the bottom of a fine
capillary tube.
• The scale is calibrated from 0 to 6 K and
subdivided into 0.01 K.
• The unique feature of this thermometer, however,
is the small reservoir of mercury at the top.
• The amount of mercury in this reservoir can be
decreased or increased by tapping the
thermometer gently.
• In this way the thermometer is adjusted so that
the level of mercury thread will rest at any
desired point on the scale when the instrument is
placed in the boiling (or freezing) solvent.
146

• Procedure. The apparatus is fitted up as shown
• Solvent is placed in the boiling tube with a porcelain piece lying
in it.
• It is heated on a small flame (micro burner).
• As the solution starts boiling, solvent vapour arising from the
porcelain piece pump the boiling liquid into the narrow stem.
• Thus a mixture of solvent vapour and boiling liquid is
continuously sprayed around the thermometer bulb.
• The temperature soon becomes constant and the boiling point of
the pure solvent is recorded.
• Now a weighed pellet of the solute is added to the solvent and
the boiling point of the solution noted as the temperature
becomes steady.
• Also, the volume of the solution in the boiling tube is noted.
• The difference of the boiling temperatures of the solvent and
solution gives the elevation of boiling point.
• While calculating the molecular weight of solute the volume of
solution is converted into mass by multiplying with density of
solvent at its boiling point
147

• Freezing point/ melting point – temperature at which solid and
liquid phases are in equilibrium under a pressure of 1 atm.
149

• When solute is added, FP < Normal FP
• FP is depressed when solute inhibits solvent from
crystallizing.
150
When solution freezes the solid form is almost always pure.
Solute particles does not fit into the crystal lattice of the
solvent because of the differences in size.
The solute essentially remains in solution and blocks other
solvent from fitting into the crystal lattice during the freezing
process.

• A situation exists similar to elevation of boiling point.
• Freezing point depression is proportional to molal concentration of
the solute
Δ Tf = Kfm
Where, Δ Tf – freezing point depression
Kf – molal depression constant or cryoscopic constant
151

Relation between Depression of Freezing-
point and Lowering of Vapour-pressure
• The vapour pressure of a pure liquid changes with temperature as shown
by the curve ABC,
• There is a sharp break at B where, in fact, the freezing-point curve
commences.
• Thus the point B corresponds to the freezing point of pure solvent, Tf.
• The vapour pressure curve of a solution (solution 1) of a non-volatile
solute in the same solvent is also shown.
• It is similar to the vapour pressure curve of the pure solvent and meets
the freezing point curve at F, indicating that T1 is the freezing point of
the solution.
• The difference of the freezing point of the pure solvent and the solution
is referred to as the Depression of freezing point.
• It is represented by the symbol ΔT or ΔTf .
• Tf – T1 = Δ T
152

• When more of the solute is added to the solution 1,
we get a more concentrated solution (solution 2.)
• The vapour pressure of solution 2 meets the
freezing-point at C, indicating a further lowering
of freezing point to T2.
• For dilute solutions FD and CE are approximately
parallel straight lines and BC is also a straight line.
• Since the triangles BDF and BEC are similar,
• where p1 and p2 are vapour pressure of solution 1
and solution 2 respectively.
• Hence depression of freezing point is directly
proportional to the lowering of vapour
pressure.
• ΔT ∝ p – ps/p
153

Determination of Molecular Weight
from Depression of Freezing point 154

• The value of Kf can be determined by measurement of ΔT by
taking a solute of known molecular mass (m) and substituting
the values in expression (6). The constant Kf , which is
characteristic of a particular solvent, can also be calculated from
the relation
• where Tf = freezing point of solvent in K; Lf = molar latent heat
of fusion; R = gas constant.
• Hence for water, Tf = 273 K and Lf = 336 J g–1. Therefore,
155

Determination of freezing
point depression
2 methods
1. Beckmann method
2. Equilibrium method
156

Beckmann’s Method (1903)
• Apparatus: It consists of
• (i) A freezing tube with a side-
arm to contain the solvent or
solution, while the solute can
be introduced through the
side-arm;
• (ii) An outer larger tube into
which is fixed the freezing
tube, the space in between
providing an air jacket which
ensures a slower and more
uniform rate of cooling;
• (iii) A large jar containing a
freezing mixture e.g., ice and
salt, and having a stirrer.
157

• Procedure. 15 to 20 g of the solvent is taken in the freezing point of
the solvent by directly cooling the freezing point tube and the
apparatus is set up as shown so that the bulb of the thermometer is
completely immersed in the solvent.
• First determine the approximate freezing point of the solvent by
directly cooling the freezing point tube in the cooling bath.
• When this has been done, melt the solvent and place the freezing-
point tube again in the freezing bath and allow the temperature to
fall.
• When it has come down to within about a degree of the approximate
freezing point determined above, dry the tube and place it cautiously
in the air jacket.
• Let the temperature fall slowly and when it has come down again to
about 0.5° below the freezing point, stir vigorously.
• This will cause the solid to separate and the temperature will rise
owing to the latent heat set free.
• Note the highest temperature reached and repeat the process to get
concordant value of freezing point.
158

• The freezing point of the solvent having been accurately
determined, the solvent is remelted by removing the tube
from the bath, and a weighed amount (0.1–0.2 g) of the
solute is introduced through the side tube.
• Now the freezing point of the solution is determined in
the same way as that of the solvent.
• Knowing the depression of the freezing point, the
molecular weight of the solute can be determined by
using the expression
159

• This method gives accurate results, if the following
precautions are observed :
(a) The supercooling should not exceed 0.5°C.
(b) The stirring should be uniform at the rate of about one
movement per second.
(c) The temperature of the cooling bath should not be 4° to
5° below the freezing point of the liquid
160

Rast’s Camphor Method
• This method due to Rast (1922) is used for determination
of molecular weights of solutes which are soluble in
molten camphor.
• The freezing point depressions are so large that an
ordinary thermometer can be used
161

• Pure camphor is powdered and introduced into a capillary tube which is
sealed at the upper end.
• This is tied along a thermometer and heated in a glycerol bath.
• The melting point of camphor is recorded.
• Then a weighed amount of solute and camphor (about 10 times as much)
are melted in test-tube with the open end sealed.
• The solution of solute in camphor is cooled in air.
• After solidification, the mixture is powdered and introduced into a
capillary tube which is sealed.
• Its melting point is recorded as before.
• The difference of the melting point of pure camphor and the mixture,
gives the depression of freezing point.
• In modern practice, electrical heating apparatus is used for a quick
determination of melting points of camphor as also the mixture.
• The molal depression constant of pure camphor is 40°C.
• But since the laboratory camphor may not be very pure, it is necessary to
find the depression constant for the particular sample of camphor used by
a preliminary experiment with a solute of known molecular weight.
162

Kb and Kf values of some solvents
164

Applications
• Salting of snow
• Antifreeze used in car batteries –
Propylene glycol, glycerol, honey etc
165
Elevation of boiling point and
depression of freezing point can help
in determining molecular weights of
substances

• Diffusion – both the solute and solvent molecules travel freely
• Osmosis – only solvent molecules pass through the semi permeable
membrane.
• Passage of solvent molecules through a semipermeable membrane, from an
area of low solute concentration to an area of high solute concentration is
called as osmosis.
167

• This happens by equalization of escaping tendency
of the solvent on both the sides of the membrane.
• Escaping tendency can be measured as osmotic
pressure.
168

Osmotic Pressure - The Pressure that must be applied
to stop osmosis
169

Preparation
• Addition of non-volatile solute to solvent
• Lowering of vapour pressure
Setup
• Pure solvent is placed adjacent to above solution but separated
by semipermeable membrane
• Solvent molecule pass through membrane into the solution to
dilute out the solute and to raise the vapour pressure back to
original value
Measurement
• Osmotic pressure is measured by:
• Measuring the hydrostatic head appearing in the solution
• Applying a known pressure which just balances the osmotic
pressure
170

Osmotic pressure: A lowering of vapour pressure
• Works on the principle of thistle tube
apparatus
• Once equilibrium is obtained, height on
the solution side of the membrane is
greater than the height on the solvent
side
П = hρg
П – osmotic pressure
h – difference in heights
ρ – solution density
g – acceleration due to gravity
171

MEASUREMENT OF OSMOTIC
PRESSURE
172

• Two methods:
• Measuring the hydrostatic head appearing in the solution
– NOT USED
• Applying a known pressure which just balances the
osmotic pressure i.e. there is no passage of solvent
molecules through the semipermeable membrane
• Pressure can be measured by a manometer or by more
sensitive electric techniques
173

van’t Hoff equation
• 1886 – Jacobus van’t Hoff recognised a relationship
between osmotic pressure, concentration and temperature
• He concluded that,
174
op in a dilute
solution
pressure that the solute
would exert if it were a
gas occupying the same
volume

• Where,
• П – osmotic pressure in atm
• V – volume of solution in liters
• n – number of moles of solute
• R – gas constant
• T – absolute temperature
175
ПV = nRT
V
nRT

Where,
c = concentration of solute in
moles/ liter (molarity)

Morse equation
• He proved that if concentration when expressed in terms
of molality rather molarity, the experimental results were
more accurate.
П = RTm
176

A cell placed in an
isotonic solution. The
net movement of water
in and out of the cell is
zero because the
concentration of
solutes inside and
outside the cell is the
same.
177

• If the solute concentration
outside the cell is greater
than that inside the cell,
the solution is hypertonic.
• Water will flow out of the
cell, and crenation results.
178
Cells
shrink

• If the solute concentration
outside the cell is less than
that inside the cell, the
solution is hypotonic.
• Water will flow into the
cell, and hemolysis results.
179
Cells
burst

• Osmotic pressure can be used in determination of the
molecular weight of substances
182

Choice of colligative property
for determination of molecular
weight
Boiling point
elevation
• Solute is non-
volatile
• Doesn’t
decompose at
bp
Freezing point
depression
• Solute should
be volatile
• High accuracy
for solutions
of small
molecules
Osmotic
pressure
• No difficulties
• Widely used
for molecular
weight
determination
of polymers
183

Reference:
• Physical Pharmacy by Martin
184

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