Solution colligative properties 12th HSC Maharashtra state board

Solution Colligative Properties
Presentedby:FreyaCardozo
1
Presented by:
Freya
Cardozo

Solution
 It is a homogenous mixture of two or more pure substances
 Solution= Solute + Solvent
 Solvent is the one present in larger quantity
 Solute is the one present in smaller quantities
 Solutions with only one solute are called Binary solutions
Presented by: Freya Cardozo
2

Types of solutions
 The solute and solvent can be in any of the following state solids, liquids
or gases
 Combinations of these gives mainly 9 types of solutions
3

Solubility
 The solubility of a solute is its amount per unit volume of
saturated solution at a specific temperature
 Solubility unit= mol L-1
5

Factors affecting solubility
 Nature of solute and solvent
 Effect of temperature on solubility
 Effect of pressure on solubility
6

Nature of solute and solvent
 Like dissolves like
 Similar chemical character are more readily soluble in each other than
different chemical natures
 Similar substances also have similar intermolecular forces of attraction
 Polar solutes dissolve in polar solvents because , solute-solute, Solute-
solvent and solvent-solvent interactions are of similar magnitude
 Eg. NaCl in water, Cholesterol in Benzene, Sugar in water
7

Effect of temperature on solubility
 For endothermic processes – Eg. KCl in water
↑T ↑Solubility because there is an increase in stress on the solution
 For exothermic process – CaCl2 in water
↑T ↓Solubility
But there hasn’t been any direct relation between exo and
endothermicity on T
8

T Vs Solubility
 NaBr, NaCl, KCl : S slightly changes with T
 KNO3, NaNO3, KBr: ↑T ↑Solubility
 Na2SO4 ↑T ↓Solubility
Gas molecules are condensed in liquid phase, this
process is exothermic. Thus, Solubility of gases in
water should decreases with increase in
temperature.
9

REAL LIFE EXAMPLE
Industries take billions of gallons of water from rivers and lakes to cool
down the equipment's which get heated during the industrial
production process. Once the cool water is used the hot water
produced is the returned to the water bodies again. Due to the
increase in the temperature of water the solubility of gas in water
decrease and thus it is very less available for fishes
10

Effect of pressure on solubility
 Pressure has no effect on solubility of liquids and solids as they are
incompressible
 But for gases the solubility is greatly influenced by pressure
 The relation between P and S for gases can be given by HENRY’S
LAW
11

HENRY’S LAW
 The law states that the solubility of a gas in a liquid is directly proportional to the
pressure of the gas over the solution

 Where, S= Solubilty of gase in molL-1
P=Pressure of gas over solution in bar
KH= Henry constant. Unit is mol L-1 bar-1
 When, P=1, KH=S. Thus, KH is the solubility of the gas in a liquid when its pressure over
solution is 1 bae
12

Example and Exceptions
 CO2 added in cold drinks under a pressure
 As compared to normal conditions under pressure the solubility is much higher
 Release relives this pressure giving the effervescence
 Exceptions are NH3 and CO2 because they react with water
 As they form these compounds they have higher solubilities than expected by
Henrys law
13

Type 1 numericals
 Based on Henry’s law
 S=KHP
14

QUESTION
 The henry’s law constant of CH3Br is 0.159 mol L-1 bar-1 at 250C.
What is the solubility of CH3Br in water at 250C and at 130mm Hg?
(1mmHg= 0.00133 bar)
15

RAOULT’S LAW
 The law states that, “ The partial vapour pressure of any volatile
component of a solution is equal to the vapour pressure of the
pure component multiplied by its mole fraction in the solution.”
 P= x P0
 P= partial pressure
 p0= partial pressure of pure component
 x= mole fraction
16

Mole fraction
 Mole fraction represents the number of molecules of a particular component in
a mixture divided by the total number of moles in the given mixture. It's a way
of expressing the concentration of a solution
 Represented by x
 The mole fraction of solute is x2 and solvent is x1; x1+x2=1
17

Derive Raoult’s law
1. Let A1 and A2 be two volatile liquids, P1 and P2 be the respective partial pressures
and P1
0 , P2
0 be vapour pressures of pure liquids
2. Let x1 and x2 be the mole fractions of A1 and A2 respectively
3. Raoult’s law can be written as P1= x1 P1
0 and P2 = x2 P2
0
18

Since, P1
0 and P2
0 are constant
P Vs x2 is a straight line
P1 Vs X1 and P2 Vs x2 are also straight line passing
through the origin
19

IDEAL SOLTIONS NONIDEAL
SOLUTIONS
1. Obey Raoult’s law over entire range of
concentrations
2. V.P always lies between V.P of pure components
3. No heat is evolved/ absorbed when 2
components are mixed. Enthalpy of mixing is
zero. ∆mix H=0
4. No volume change. Thus Vol of sol. Is equal to
sums of volumes of the 2 components mixed.
∆mixV=0
5. Solvent-solute, solute-solute and solvent-solvent
molecular interactions are comparable
1. Do not obey Raoult’s law over entire range
of concentrations
2. V.P of these solutions can be higher or lower
than those of pure components
3. It shows two types of deviation:
 Positive deviation
 Negative deviation
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POSITIVE DEVIATION NEGATIVE DEVIATION
 Solute-solvent I.F.A (weaker)< solute-solute
I.F.A & solvent-solvent I.F.A
 V.P is higher than the pure components
 Eg. Acetone+ Ethanol
Acetone+ carbon disulphide
 Solute-solvent I.F.A (stronger)> solute-solute I.F.A
& solvent-solvent I.F.A
 V.P is higher than the pure components
 Eg. Acetone + chloroform
Phenol + Aniline
21
I.F.A= Intermolecular forces of attraction

Colligative properties
 Definition of colligative properties
The physical properties of solutions that depend on the number of solute particles in
solutions and not on their nature are called colligative properties
 There are 4 colligative properties
1. Vapour pressure lowering
2. Boiling point elevation
3. Freezing point depression
4. osmotic pressure
 Non-electrolytic and dilute solutions are used
22

Vapor pressure
 The pressure exerted by the liquid vapor over the surface of a liquid in a closed
container when both are in equilibrium is called as vapor pressure
23

VOLATILE SUBSTANCE NON-VOLATILE SUBSTANCE
24

Vapour pressure lowering
 When a non volatile non-ionizable solid is dissolved in a liquid solvent, the vapour pressure of the
solution is lower than that of the solvent.
 If the solute is non-volatile it does not contribute to the VP above the solution
 Thus VP of solution=VP of solvent above the solution
 Mathematically, If P1
0 is the VP of pure solvent and P1 is the VP of solvent above the solution, P1<
P1
0
 Thus, vapour pressure lowering is ∆= P1
0 – P1
25

Reason for this?
 The vapour pressure of a liquid depends on the ease
with which the molecules escape from the surface of
liquid
 When non-volatile solute is added it replaces some of
the volatile solvent molecules at the surface and these
non volatile solutes do not vaporizes and thus do not
contribute to the vapour pressure.
 Therefore, the number of solvent molecules available for
vaporization per unit surface area is less than number at
surface of pure solvent
26

Raoult’s law for solution of non volatile solutes OR Prove that
∆P= P1
0 x2 OR Prove lowering of vapour pressure is a
colligative property
 For a solution containing non volatile solute, the vapour pressure of
solvent over the solution is equal to the vapour pressure multiplied by its
mole fraction in the solution
P1= P1
0 x1
 For a binary solution containing 1 solute, x1=1-x2 (Since, x1+x2=1)
 But, we know that lowering of vapour pressure is given by ∆= P1
0 – P1
 Thus, ∆P= P1
0x2
 From the above equation it is clear that the ∆P depends on x2 which is
number of solute particles, thus lowering of vapour pressure is a colligative
property
27

Relative Lowering of Vapour pressure
 The ratio of vapour pressure lowering of solvent divided by the vapour pressure of pure solvent is
called relative lowering of vapour pressure.
 Thus,
 Relative lowering of VP is equal to the mole fraction of solute in the solution. Therefore, relative
lowering of VP is also a colligative property
28

Relationship between molar mass of
solute and lowering of vapour pressure
 We know that relative lowering of vapour pressure is equal to mole fraction of solute i.e.
 The mole fraction of a component of solution is equal to its moles divided by the total moles
in the solution. Thus,
 n1= moles of solvent and n2= moles of solute
 In dilute solutions, n1>>n2, thus n1+n2~n1.
 Thus the mole fraction is
29

 Suppose a solution is prepared by adding W2 g of solute in W1 g of solvent. The moles of
solute and solvent in the solution are,
 Where, M1 and M2 are molar masses of solvent and solute respectively. Substituting in the
equation,
30

WHAT DO YOU THINK?
31

TYPE 2 numericals
∆P= P1
0X2
32

33In an experiment, 18.04 g of mannitol was dissolved in 100 g of water. The vapour pressure of
water was lowered by 0.309 mm Hg from 17.535 mmHg. Calculate the molar mass of
mannitol

34
A solution is prepared by dissolving 394g of a non volatile solute is 622g of water. The vapour
pressure of solution is found be 30.74 mm Hg at 300C. If the vapour pressure at 300C is 31.8
mm Hg, what is the molar mass of solute?

35The vapour pressure of pure benzene (molar mass 78 g/mol) at a certain temperature is 640
mmHg. A non-volatile solute of mass 2.315g is added to 49g of benzene. Th vapour pressure
of solution is 600 mmHg. What is the molar mass of the solute?

36

Boiling point elevation
 B.P : The temperature at which the vapour pressure equals to the applied
pressure/ atmospheric pressure(if open container)
 Solutions with non volatile solute have higher B.P than pure solvent
 If Tb
0 is B.P of pure solvent and Tb is of the solution then the difference between
them ∆Tb
 The difference between the B.P of solution and that of the pure solvent at
any given point is called the boiling point elevation
37

Lowering of V.P and Elevation in B.P
1. In the graph Vapour pressure of the solvent and solution are
plotted as a function of temperature
2. From the previous discussions we know that the vapour pressure of
solution with non volatile solute<vapour pressure of pure solvent
3. Thus from graph we can see that
CD= vapour pressure of solution(lower curve)
AB= vapour pressure of pure solvent(higher curve)
4. The V.P difference goes on increasing with temperature
38

5. The intersection of the curves AB and CD on the X axis shows that
the boiling point of solution is more than the pure solvent
6. A liquid boils when its V.P is equal to 1 atm(atmospheric
pressure). Therefore, in order for the solution to boil a higher
temperature will be needed to reach 1 tm than the solvent.
7. That’s why the solution needs to be heated more to be boiled.
Hence, there will be elevation in the B.P of solution even though
there is a lowering of vapour pressure.
[V.P is inversely related to B.P]
39

Expressing concentration
MOLARITY MOLALITY
40

B.P elevation and conc. Of solute
1. The B.P elevation is directly proportional to molality of the solution.
2. Where, m= molality of solution
Kb= boiling point elevation constant/ molal elevation constant/ ebullioscopic constant
3. If, m=1, ∆Tb=Kb
Thus, ebullioscopic constant is the B.P elevation produced by 1 molal solution
41

Batao..
42

Why molality over molarity?
Because, we are studying systems where temperature is not constant thus we have to
choose terms that do not depend on temperature
molality is temperature independent whereas molarity depends on temperature
This can be seen from the units also
43
Molality= mol/kg
Molarity= mol/L

Relationship between mass of solute and
boiling point elevation
 Suppose a solution is made by dissolving W2 g of solute in W1 g of solvent
 Moles of solute in solvent= W2/ M2 (M2= molar mass)
Mass of solvent= W1g = W1 g/ 1000 g/kg = W1/ 1000 kg
 Molality can be given by
44

TYPE 3:
Elevation of B.P
45

46

47

48

49
A solution containing 0.73 g of camphor (molar mass 152 gmol-1) in 36.8 g of acetone
(boiling point 56.30C) boils at 56.55 0C. A solution of 0.564 g of unknown compound in the
same weight of acetone boils at 56.46 0C. Calculate the molar mass of the unknown
compound. [Oct 2014]

Freezing point
 Freezing of point of a liquid is the temperature at which the liquid and solid are
in equilibrium and the two phases have the same vapour pressure
50

Depression in Freezing point
 It is observed that addition of non volatile solute to solvent lowers the freezing
point
 Thus, The F.P of solution( with non-volatile solute)< F.P of pure solvent
 Now, if Tf
0 is freezing point of pure solvent and Tf of the solution, Tf
0 > Tf
 The difference between the two is called ∆ Tf
51

Freezing point depression & Vapour
pressure lowering
 Consider the vapour pressure diagram
1. Curve AB= V.P of solid solvent
2. Curve CD= V.P of pure liquid
3. Curve EF = V.P of solution
 The vapour pressure of solution is always lower than the
solvent and thus is present at the lower part of the graph
 The non-volatile solute does not dissolve in the solid solvent
52

 Curve AB and CD  Intersect at B – This is where both the
solid and liquid phases have the same V.P.
Thus, The temperature corresponding to point B is freezing
point of the pure solvent Tf
0
 Curve EF and AB  Intersect at E – This is the point at
which the solid solvent is in equilibrium with the solution
Thus, The temperature corresponding to point E is freezing
point of the solution Tf
 It is clear from the figure that the freezing point of solution
is lower than that of the pure solvent
53

But why so?
DUE TO THE ATTRACTIVE FORCES BETWEEN THE MOLECULES
 In pure liquid the attractive forces among molecules are large enough to cause
the change of phase from liquid to solid
 In solution, there are solute molecules in between the solvent particles.. this
causes more separation of solvent molecules than in the solvent
 Therefore, there is a decrease in the attractive forces between the solvent
molecules. So the temperature of the solution is lowered below the freezing point
of solvent to cause phase channge
54

55
Only solvent molecules Solute + solvent molecules
Easier to change
phase
Solute molecules obstruct the
phase change. More T
needed(Lower Temps)

Real life application
56

Freezing point depression and molality
 Freezing point depression ∆Tf is directly proportional to molality of the solution
 Kf= Freezing point depression constant / cryoscopic constant
 If, m=1 ; ∆Tf = Kf
 The cryoscopic constant thus is the depression in freezing point produced by 1 molal solution
of a non volatile solute
57

58

Relationship between Molar mass of
solute and F.P depression
59

60

61

Osmosis
 The net flow of solvent molecules into the solution through a semipermeable membrane
OR
 The net flow of solvent molecules into the solution from a more dilute solution to more
concentrated solution through a semipermeable membrane is called osmosis
62
SOLVENT SOLUTION Lower Conc.
Higher
conc.

Semipermeable membrane
 It is a film such as cellophane which has pores large
enough to allow the solvent molecules to pass through
them
 The pores are small enough to not allow the flow of the
larger solute molecules or ions of higher mass
 Selectively allows passage of solvent molecules
63
Solute
molecules
Semi-
P.M
Solvent
molecules

Osmotic pressure
 In the thistle tube solution of interest (sugar solution)
is placed and it is immersed in beaker filled with pure
water
 A semipermeable membrane is placed at the mouth
of the tube
 Some solvent passes through the membrane into the
solution
 This causes rise in liquid level in the tube
 Now, the hydrostatic pressure in the tube pushes the
solvent back into the container
64

Reverse osmosis
 The liquid rises in tube and then stops, when the liquid stops
rising that is the pressure which stops the flow and is called
the osmotic pressure
 This hydrostatic pressure that stops the osmosis is of the
solution is the osmotic pressureΠ and is equal to
1. Height of the liquid column
2. Density of liquid column
3. acceleration due to gravity
65

66
Type Definition Examples
Isotonic solutions Two or more solutions having the
same osmotic pressure are said to
be isotonic
No net flow of solvent in either
direction
0.1 M urea= 0.1 M sucrose
Both have equal osmotic pressure
but
Different conc in g/L
Hypertonic solutions If two solutions have unequal
osmotic pressure, the more
concentrated solution with the
higher osmotic pressure is said to
be hypertonic
In a sucrose AND urea solution,
Sucrose Higher osmotic
pressureHypertonic
Hypotonic solutions If two solutions have unequal
osmotic pressure, the more dilute
solution with the lower osmotic
pressure is said to be hypertonic
In a sucrose AND urea solution,
Urea Lower osmotic
pressureHyptonic

67

Osmotic pressure and conc. Of solution
 For dilute solution the osmotic pressure can be given
by,
 V= volume of the solution in dm3
n2= number of moles of non volatile solute
R= real gas constant = 0.08206 dm3 atm K-1 mol-1
Π= Osmotic pressure in atm
68
 We know that Concentration= n2/V
 Concentration can be written in terms of
Molarity M
 Thus, equation becomes
 Here, we can use molarity instead of
molality because the osmotic pressures
are measured at a constant temperature.

69

Molar mass of solute from osmotic pressure
70

Reverse osmosis
 The direction of osmosis if from pure solvent to solution but
this can be reversed by applying a pressure higher than the
osmotic pressure
 The pure solvent then flows from solution into pure solvent
through semipermeable membrane. This is called reverse
osmosis.
 Eg. Fresh water and salty water separation using semipermeable
membrane
 On application of pressure higher than osmotic pressure, the
salty water passes into fresh pure water
 This leaves the salt behind
71

72

Colligative properties of electrolytes
1. The solutions of electrolytes also exhibit colligative properties which do not obey
the relations of non-electrolytes
2. The colligative properties of the solutions of electrolytes are greater than those to
be expected for solutions of non electrolytes of the same concentrations
3. The molar masses of electrolytes in aqueous solutions determined by colligative
properties are found to be considerably lower than the formula masses
73

Why is colligative properties of
electrolytes more than nonelectrolytes
Electrolytes-
Dissociates
Increases the
number of
particles
Increase in
colligative
particles
Eg. NaCl(lesser
C.P) Vs
Sucrose(more
C.P)
74

Van’t Hoff factor (i)
 In order to account for the dissociation/ association of electrolytes and to calculate their colligative
properties Van’t Hoff suggested the factor i
 It can be defined as the ratio of colligative property of a solution of electrolyte divided by the
colligative property of nonelectrolyte solution of the same concentration.
 No subscript  electrolyte solutions With subscript nonelectrolyte solutions
75

Definitions
 It is also defined as ,
76
i=2
KNO3
NaCl
i=3
CaCl2
Na2SO4
i=1 for non
electrolytic
solutions

Why colligative properties of higher
concentration solutions smaller than
expected?
 The electrostatic forces between the oppositely charged ions bring about the
formation of ion pairs
 Each ion pair consists of one or more cations and one or more anions held
together by electrostatic attractive forces
 This results in decrease in the number of particles in solution causing reduction in
the expected I value and colligative properties
77

Modifications of expressions of colligative
properties
78

Van’t Hoff factor i and degree of dissociation
 For weak electrolytes Dissociation is related tαo Degree of dissociation α
 Consider the equation
 Initially 1 mol 0 0
At equilibrium (1- α)mol x α y α
Total moles after dissociation = (1- α)+ x α + y α= 1+ α (x+y-1)= 1+ α(n-1) [Since, n=x+y]
Vant Hoff factor can thus be given by,
79

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Solution colligative properties 12th HSC Maharashtra state board

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