This document discusses colligative properties of solutions and ways of expressing concentration. It begins by defining key terms like solute, solvent, concentration, dilute and concentrated solutions. It then describes various ways of expressing concentration including percentage by weight, mole fraction, molarity, molality, normality and parts per million. The document also discusses colligative properties like lowering of vapor pressure, elevation of boiling point, depression of freezing point and osmotic pressure. It provides equations and experimental methods for determining these properties and using them to calculate molecular masses. The concept of abnormal molar masses from association or dissociation in solution is introduced along with the van't Hoff factor.

1. Colligative Properties of Solution
B.Sc. SEM-V
Paper-II (Physical Chemistry)
Dr. N. G. Telkapalliwar
Associate Professor
Department of Chemistry
Dr. Ambedkar College, Nagpur

2. Solution: A solution is a homogeneous mixture of two or more
substances.
 Solute
 Solvent
Concentration of the solution: It is defined as the amount of solute
present in a given amount of solution.
Concentration is generally expressed as the quantity of solute in a
unit volume of solution.
 Dilute solution
 Concentrated solution.
Introduction

3. WAYS OF EXPRESSING CONCENTRATION
Percentage by weight
 Mole fraction
 Strength
 Molarity
 Molality
 Normality
 Parts per million (ppm)

4. 1) Percentage by Weight
It is the weight of the solute as a per cent of the total weight of the solution.
Example:
If a solution of HCl contains 36 per cent HCl by weight, it has 36 g of HCl for 100 g of
solution.

5. 2) Mole fraction
A simple solution is made of two substances : one is the solute and the other solvent.
Mole fraction (X) of solute is defined as the ratio of the number of moles of solute
and the total number of moles of solute and solvent.
Thus,

7. 3) Strength of Solution
Strength of a solution: It is defined as the number of grams of solute present
per liter of the solution.
Molarity (M) is defined as the number of moles of solute per litre of
solution.
4) Molarity

9. 5) Molality
Molality (m) of a solution is defined as the number of moles of solute dissolved in one
kilogram of the solvent.

10. 6) Normality
Normality (N) is defined as the number of gram equivalent of a solute present per
litre of solution.
If 40 g of NaOH (eq. wt. = 40) be dissolved in one litre of solution, normality of
the solution is one and the solution is called 1N (one-normal).
A solution containing 4.0 g of NaOH is 1/10 N or 0.1 N or decinormal.

12. 7) Parts per million (ppm)
Parts per million (ppm) : It is defined as the number of parts of the solute present in
one million parts of solution.
In case of solution of solid in liquid,
“It is defined as the number of milligrams of a solid solute present in one liter of
solution”.
Ex: A 350 mL sample of drinking water was analyzed and found to contain 0.0046 g of
sulphate salts. Calculate the concentration of sulfate salts in this water sample?

13. Colligative Properties
The properties which depends on the number of particles in solution and
not on the nature of the particles are called as colligative properties.
Dilute solutions containing non-volatile solute exhibit the following
colligative properties :
(1) Lowering of the Vapour Pressure
(2) Elevation of the Boiling Point
(3) Depression of the Freezing Point
(4) Osmotic Pressure

14. Compare the properties of 1.0 M Sucrose solution to a
0.5 M solution of NaCl.
 Despite the conc. of NaCl = ½ the conc. of sucrose both solutions
have precisely the same number of dissolved particles, why?
 Because each NaCl unit creates two particles upon dissolution a
Na+ and a Cl-.
 Both solutions have the same freezing point, boiling point, vapor
pressure, and osmotic pressure because those colligative properties
of a solution only depend on the number of dissolved particles.
 Other non-colligative properties including e.g. viscosity, surface
tension, and solubility are different.

15. Lowering of the Vapour Pressure (Raoults’s Law)
Raoult’s law: The partial pressure of any volatile component of
a solution at any temperature is equal to the product of vapour
pressure of the pure component and mole fraction of that
component in the solution.
Ps = XA . P
Ps = XB . P
Where,
Ps = Vapour pressure of a solution
P = Vapour pressure of pure solvent
XA= Mole fraction of solute on solution
XB = Mole fraction of solute on solution

16. The vapour pressure of a pure solvent is decreased when a non-
volatile solute is dissolved in it.
If p is the vapour pressure of the solvent and ps is the vapour pressure
of the solution, then lowering of vapour pressure is (p – ps).
This lowering of vapour pressure relative to the vapour pressure of
the pure solvent is termed the Relative lowering of Vapour pressure.
Thus,
Lowering of the Vapour Pressure

17. Raoults’s law: The relative lowering of the vapour pressure of a
dilute solution is equal to the mole fraction of the solute present
in dilute solution.
Raoult’s Law can be expressed mathematically in the form :
Where,
n = number of moles or molecules of solute
N = number of moles or molecules of solvent.

18. The vapour pressure of the pure solvent is caused by the number of molecules
evaporating from its surface.
When a nonvolatile solute is dissolved in solution, the presence of solute molecules in
the surface blocks a fraction of the surface where no evaporation can take place.
Derivation of relative lowering of the Vapour Pressure (Raoult’s Law)
This causes the lowering of the vapour pressure. The vapour pressure of the solution
is, therefore, determined by the number of molecules of the solvent present at any
time in the surface which is proportional to the mole fraction.
That is,
Where, N = moles of solvent and n = moles of solute.
Where, k being proportionality constant .
---------(1)

21. Determination of Molecular Mass

24. Experimental determination of lowering of vapour pressure
Ostwald and Walker’s Dynamic Method (Gas Saturation Method)
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.

25. Calculations:
As the air bubbles through set A it is saturated up to the vapour pressure ps of solution
and then up to vapour pressure p of solvent in set B. Thus the amount of solvent taken
up in set A is proportional to ps and the amount taken up in set B is proportional to
(p – ps).
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).

26. Osmosis and osmotic pressure

27. Let us consider a pure solvent and solution separated by a membrane
which permits the passage to solvent molecules but not to solute molecules.
Only the solvent will diffuse through the membrane into solution.
A membrane which is permeable to solvent and not to solute, is called a
semipermeable membrane.
The net flow of the solvent molecules from lower concentration to higher
concentration region through semipermeable membrane is called as
Osmosis.
WHAT IS OSMOSIS?

28. What is osmotic pressure?
The minimum pressure to be applied on the solution to stop the process of
osmosis is called as osmotic pressure.
Or
Osmotic pressure may be defined as the external pressure applied to the
solution in order to stop the osmosis of solvent into solution separated by
a semipermeable membrane.
External pressure

29. What is Reverse Osmosis?
 Reverse Osmosis works on the same principle as osmosis, but in the reverse
direction.
 In this Reverse osmosis process direction of the flow of the solvent
molecules are reversed by applying greater pressure.
The flow of the solvent molecules from higher concentration to lower
concentration region through semipermeable membrane under high pressure is
called reverse osmosis.
Osmosis Reverse Osmosis

30. Measurement of Osmotic pressure
Berkeley and Hartley’s Method
Berkeley and Hartley (1904-1909) employed the technique of applying external
pressure on the solution just enough to prevent osmosis.
Procedure:
 The inner porcelain tube is filled with pure solvent and the jacket with the solution
whose osmotic pressure is to be determined.
 The level of the solvent meniscus in the capillary tube will tend to move down as
solvent flows into the solution across the membrane.
 Pressure is then applied through the piston so that the meniscus becomes stationary.
It indicates that osmosis has been stopped and now the pressure recorded by the
pressure gauge gives the osmotic pressure of the solution.
Advantages
(a) It is quick and accurate.
(b) It can be used for determining high
osmotic pressures.
(c)The osmotic pressure being balanced
by the external pressure, there is no strain
left on the membrane and the danger of its
bursting is eliminated.

31. Determination of molecular mass from osmotic pressure
The method of determination of molar mass on the basis of osmotic
pressure is based on vant Hoff’s two laws:-
1) Boyle-vant Hoff’s law : At constant temperature, osmotic pressure of a
dilute solution is directly proportional to molar concentration of the
solution.
2) Charle-vant Hoff’s law: For a given concentration of a solution, the
osmotic pressure is directly proportional to the absolute temperature.
That is, π ∝ C at constant temperature, Where π is osmotic pressure
That is, π ∝ T at constant concentration, Where π is osmotic pressure

32. By combining the two laws we get,
π ∝ C T or π = R C T
But Concentration is expressed as number of moles of solute per unit
volume of the solution.
That is, C = n/V
Therefore, π = nRT /V
πV = nRT
πV = wRT/M
M = wRT/ πV
Where,
M = molecular mass of the solute
w = amount of solute in grams
R = 0.0821 litre-atmosphere
T = (t°C + 273) K
π = osmotic pressure in atmospheres
V = volume of solution in litres

35. Elevation of boiling point
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
The vapour pressure curves of the
pure solvent, and solutions-1 and 2
with different concentrations of
solute are shown in Figure.

36. Ostwald-Walker method of measuring the relative lowering of vapour pressure.
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 ...(1)

37. Determination of Molecular Mass from Elevation of Boiling Point
The elevation of boiling point is directly proportional to the lowering of vapour pressure.
ΔT ∝ p – ps ...(1)
Since p is constant for the same solvent at a fixed temperature, from (1) we can write
But from Raoult’s Law for dilute solutions,
Since M (mol mass of solvent) is constant, from (3)
From (2) and (4)

38. 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.

39. The constant Kb, which is characteristic of a particular solvent used, can
also be calculated from thermodynamically derived relationship.
Where,
R = gas constant;
Tb = boiling point of solvent;
Lv = molar latent heat of vaporization.

42. Depression of freezing point
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 nonvolatile
solute in the same solvent is also
shown in Fig. 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 = ΔT

43. Relation between lowering of vapour-pressure and depression of freezing point
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.
or ΔT ∝ p – ps ...(1)

44. Determination of Molecular Weight from Depression of Freezing point
Depression of freezing point is directly proportional to the lowering of vapour pressure.
or ΔT ∝ p – ps ...(1)
Since p is constant for the same solvent at a fixed temperature, from (1) we can write
But from Raoult’s Law for dilute solutions,
Since M (mol wt) of solvent is constant, from (3)
from (2) and (4)

45. where Kf is a constant called Freezing-point constant or Cryoscopic constant
or Molal depression constant. If w/m = 1 and W = 1, Kf = ΔT.
Thus, Molal depression constant may be defined as the freezing-point
depression produced when 1 mole of solute is dissolved in one kg (1000 g) of
the solvent.
If the mass of solvent (W) is given in grams, it has to be converted into
kilograms. Thus the expression (5) assumes the form;
Where, m = molecular mass of solute ; Kf = molal depression constant ; w = mass
of solute; ΔT = depression of freezing point ; W = mass of solvent.

46. 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.

49. Abnormal Molar Masses and Van’t Hoff Factor
Colligative properties depends upon the number of particles (molecules or
ions) of the solute present in the solution. If the solute undergoes dissociation
or association in the solution, the molar mass obtained is abnormal.
Van’t Hoff introduced a correction factor ‘i’ which is defined as follows;
Since colligative properties are inversely proportional to the molecular mass
of solute,
Such abnormal molar masses are due to two reasons
1) Association
2) Dissociation

50. Degree of association:
Degree of association means the fraction of the total number of molecules which
combine to form bigger molecules.
Consider one mole of a solute dissolved in a given volume of a solvent.
Suppose, n simple molecules combine to form an associated molecule,
i.e., nA (A)n
Let,  be the degree of association
The number of unassociated moles = 1-
The number of associated moles = /n
Therefore, The number of effective moles = 1- + /n
Since, Colligative effect is proportional to the number of moles, therefore, van’t Hoff
factor ‘i’is given by,
Thus, knowing the values of ‘i’ and ‘n’ the degree of association can be calculated

51. Degree of Dissociation:
Degree of dissociation means the fraction of the total number of molecules which
dissociates, that is, break into simple molecules or ions.
Consider one mole of uni-univalent electrolyte (KCl) dissolved in a given volume of
solvent (water).
Suppose, n simple molecules combine to form an associated molecule,
i.e., A nP
Let,  be the degree of dissociation
The number of undissociated moles = 1-
The number of dissociated moles = n
Therefore, The number of effective moles = 1- + n
Since, Colligative effect is proportional to the number of moles, therefore, van’t Hoff
factor ‘i’is given by,
Thus, knowing the values of ‘i’ and ‘n’ the degree of dissociation can be calculated

52. References
• Essentials of Physical Chemistry, S. Chand Publication by Arun Bahl, B.S. Bahl, G. D. Tuli.
• University Chemistry, Volume I, Alliance & Co., by R. M. Jugade.
• A Text Book of Chemistry, B.sc. SEM- I, Himalaya Publishing House.
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