SOLUTIONS

Solution is the homogeneous mixture of two or more non-reacting substances.

Binary solution: The solution in which one solute and one solvent is present is called binary
solution.

On the basis of physical state of solute and solvent, solutions are of 9 - types.

SOLUTE SOLVENT SOLUTION (EXAMPLE)

Solid Alloy, Amalgam
Liquid Sugar sol, Salt solution
Solid
Gas Iodine in air
Solid Hydrated salt ( CuSO4.5H2O)
Liquid dil acid, ethyl alcohol solution
Liquid
Gas Water vapour
Solid H2/Pt, Pd, Ni
Liquid Soft drinks
Gas
Gas Air

Concentration : It can be defined as the amount of solute present in given amount of solvent at a
particular temperature.

Concentrations of a solution can be expressed as follows:

(I) Strength.

(II) Mass by mass%.(w/w)

(III) Mass by volume%.(w/v)

(IV) Volume by volume%.(v/v)

(V) Mole fraction.(  )

(VI) Molality.(m)

(VII) Molarity.(M)

(VIII) Normality.(N)

(IX) Formality

(X) PPM (parts per million)

(I) Strength:

It is defined as amount of solute in gram present per litre of solution.

SI unit – g / liter.

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(II) Mass by mass%:(W/W)

mass of solute
Mass / mass% = mass of solution x 100

It has no unit.

(III) Mass by volume%:(W/V)

mass of solute
(W/V)% = volume of solution x 100.

(IV) Volume by volume%: (V/V)

volume of solute
(V/V)% = volume of solution x 100.

Eg.1. 20% (w/w) glucose solution, it means that 20 g of Glucose present in 80 g of water or
100 g of solution.

Eg.2. 95% (v/v) ethanol solution, it means that 95 ml of ethanol present in 5 ml water or
100 ml of solution.

(V) Mole Fraction: (  )

It is defined as the ratio of no. of moles of any component to the total no. of moles of
solution.

Consider a binary solution having two components 1 and 2 having no. of moles ‘n1 ’, and ‘n2
’.(1  solvent, 2  solute)

n1
x1 = n1  n2 .

n2
x2 = n1  n2 .

It has no unit.

x1 + x2 = 1.

ni
xi = n1  n2  n3 ... ni .

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 ni
n

�n i
xi = i 1 (summation over i varies from 1 to n)

(VI) Molality – (m):

It is defined as no. of moles of solute present per kg of solvent.

no. of moles of solute

Molality = mass of solvent in kg .

w2 1000
�
m = m2 w1in grms .

Where w2 = mass of solute.

m2 = molar mass of solute.

w1= mass of solvent.

Unit – mol / kg = mol kg-1 or molal – (m).

(VII) Molarity – (M):

It is defined as no. of moles of solute present per litre of solution.

no. of moles of solute

Molarity = volume of solution in litre .

w2 1000
�
M = m2 v in ml .

V = volume of solution.

Unit – mol/litre or mol lit-1 or M (Molar).

(VIII) Normality – (N):

It is defined as the no. of gram equivalent of solute present per litre of solution.

no. of gram equivalent of solute

Normality = volume of solution in lit .

w2 1000
�
N = Equivalent mass vol (lit ) .

Unit = g eq lit-1 or Normal.

Relation between N and M:

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 N = M x V.f.

Formula for Dilution:

M1V1 = M2V2.

N1V1 = N2V2.

Formula for Mixture:

M1V1 + M2V2 = M(V1 + V2)

N1V1 + N2V2. = N(V1 + V2)

(If both are either acid or Base).

If one is acid and other is Base.

N1V1 – N2V2. = N(V1 + V2)

Greater NV is taken as 1.

PPM – (Parts per million):

no.ofparts of solute
PPM of solute = totalparts of solution x 106.

Note-

(i) Between molarity and molality, molality is the best way to express concentration because it is
independent of temperature.

(ii) Between 1M and 1m, 1M is more concentrated because it has less amount of solvent.

Question- Calculate mole fraction, molality , Molarity and Normality of 20% of w/w KI solution,
the density of the solution is 1.202g/ml ( I= 127g).

Question- Calculate the mass of urea required in making 2.5Kg of 0.25 molal aq. solution.

Question- Calculate molarity of the solution when 30 ml of 0.5 M H2SO4 diluted to 500 ml.

 Solubility

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 It is defined as the maximum amount of solute which can be dissolved in 100grm of solvent at
particular temp. to make saturated solution.

On the basis of solubility solutions are of three types.

1. Unsaturated solution
2. Saturated solution
3. Supersaturated solution
 Solubility of Solid in Liquid
Factors affecting solubility
Solubility depends upon the following factors
a) Temperature
b) Pressure
c) Nature of solute and solvent
(a) Temperature: Generally increase in temperature, solubility increases because the inter molecular
space of solvent molecule increases.
�� �
Case - 1: solute + solvent �� � solution , ∆H = +ve
For endothermic solution, increase in temperature solubility increases.
Ex- NaNO3, KNO3 , NaCl , KCl etc
�� �
Case - 2 : solute + solvent �� � solution , ∆H = -ve
For exothermic solution, increase in temperature solubility decreases. ( According to Le -
Chatelier's Principle)
Ex- Li2CO3 , Na2CO3 , CeSO4 etc.
(b) Pressure : It has no effect on solubility when solute is solid and solvent is liquid.
(c) Nature of solute and solvent:
 When solute is ionic, the solubility increases in polar solvent and decrease in non polar
solvent.
 When solute is non-polar, the solubility increases in non- polar solvent.
 For ionic solute lesser is the value of lattice enthalpy, greater is the value of hydration
enthalpy, greater is the solubility.
 Solubility of gas in Liquid
Factors affecting solubility
Solubility depends upon the following factors
a) Temperature
b) Pressure
c) Nature of gas and solvent

(a) Temperature: When temperature increase solubility of gas decreases because on heating the
solution of a gas, some gas is usually expelled out from the solution.
(b) Pressure : ( Henry's Law)
Generally increase in pressure solubility of a gas in a liquid at a particular temperature increases.
 The law states that the mass of a gas dissolved in a given volume of the liquid at constant
temperature is directly proportional to the pressure of the gas present in equilibrium with the
liquid.
Mathematically m P
m = kP
 The law can also be stated as the solubility of a gas in a liquid at a particular temperature is
directly proportional to the pressure of the gas in equilibrium with liquid.

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  The law states that the partial pressure of a gas in vapour phase is directly proportional to the
mole fraction of the gas in the solution.
Mathematically
P   --------- (i)
P = KH 
Where as KH is called Henry constant.

n2
n  n2
P = KH x 1

If n2 << n1

n2
n
P = KH x 1

w2 �m1
m �w1
P = KH x 2

From equation (i)

slope = KH

 Characteristics of KH :
 Greater is the value of KH, lower is the solubility of gas at same partial pressure.
 The value of KH increases with increase in temperature.

Question- Between H2 and He, at same pressure H is more soluble in water, which gas has
greater KH value?
Ans- Helium

 Applications of Henry's Law

1. In the production of carbonated beverages - To increase the solubility of CO2 in
soft drink, soda water, beer as well as champagne, the bottles are sealed under high
pressure.
2. In the deep sea diving when the scuba diver enter in to the deep sea level, the
pressure increases hence the solubility of N2 and O2 increases in the blood. O2 gas is
used in the metabolic process where as N2 does not. When the scuba diver comes to
water surface, the pressure decreases and dissolved N2 gas produces bubble in the
blood capillary. This creates a medical condition which is known as bend. To avoid

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 bend the high conc. of N2 can be decreased by diluted with air(He). N2 = 56.2%, O2 =
32.1%, He = 11.7%
3. Anoxia - At high altitude the partial pressure of O2 is less than that of the ground
level, as a result there is low conc. of O2 in the blood and people living at high
altitude feel weak and cannot think properly. This is called anoxia.
 Limitations of Henry's Law
 The pressure should be low and the temperature should be high that means the gas
behave like an ideal gas.
 The gas should not undergo compound formation with the solvent molecules.

Question - If N2 is bubbled through water at 293K, How many milimoles of N2 gas would dissolve
in 1lit of water? Assume that N2 exerts a partial pressure of 0.987 bar.( KH for N2 =
76.48Kbar)

(c) Nature of the gas and solvent:

 Gases like H2, O2 , N2 dissolve in water to a small extent.

 Gases like H2S , NH3 , HCl are highly soluble in water.

Vapour Pressure of a liquid solution

Vapour Pressure of a solution is defined as the pressure exerted by the vapour in equilibrium
with the liquid and solution at a particular temperature.
Vapour pressure of a liquid depends upon the following factors
(i) Nature of the liquid
(ii) Temperature
 Nature of the liquid: Weaker is the intermolecular force of attraction, greater is the vapour
pressure. Ex- The vapour pressure of diethyl ether is greater than that of ethyl alcohol
 Temperature: When temperature increases, vapour pressure increases. The quantitative
relation is
�P � H vap �T2  T1 �
log � 2 �
�P1 � 2.303R � �T1 �T2 �
�
This is called Clausious- Claperon equation.
Vapour Pressure of liquid - liquid solution ( Raoult's Law)
Consider a binary mixture having two components 1 & 2 and having mole fraction 1and  2 .
When the solution is taken in a closed vessel both the components would evaporate and exist in
equilibrium.
Raoult’s Law ( For two volatile liquid)
For the solution consisting of two miscible and volatile liquids the partial vapour pressure of each
component is directly proportional to its own mole fraction in the solution at particular temperature.
Mathematically
P11
P1  P1 1 -------------(i)
o

P2 2
P2  P2  2 -------------(ii)
o
Similarly
According to Daltons law of partial pressure
PT = P1 + P2
Hence PT = P1OX1 + P2OX2
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 PT = P1OX1 + P2O(1- X1)
PT = P1OX1 + P2O - P2oX1
PT = X1 (P1O - P2O) + P2o
P1 P2
Y1 = PT and Y2 = PT
Where P1o = V.P. of component -1 in it's pure form.
P2o = V.P. of component -2 in it's pure form.
P1 = V.P. of component -1 in solution
P2 = V.P. of component -2 in solution
X1 = mole fraction of component -1 in solution
X2 = mole fraction of component -2 in solution
Y1 = mole fraction of component -1 in Vapour phase
Y2 = mole fraction of component -2 in Vapour phase
PT = total vapour pressure
Graphical representation

P20
PT = P1 + P2
v.p.
P10 P2
P1

X1 = 0 X X1 = 1
X2 = 1 X2 = 0

Question : V.P. of chloform( CHCl3) and dichloro methane( CH2Cl2) at 298K are 200mm and 415
mm Hg respectively. Calculate the V.P. of the solution prepared by mixing 25.5g of chloform and 40
g of dichloro methane at 298K. Calculate the total pressure and mole fraction of each component in
vapour phase.
Ans-
 Raoult's Law as a special case of Henry's Law

According to Henry's law P = KH
According to Raoult's law P1 = P1o X1
If P1o = KH , Raoult's law be a special case of Henry' law
 V.P. of solution of solid in liquid:
When a non volatile solute is added to the solvent, the solute particle occupies in the inter
molecular space of the solvent molecule. Hence the escaping tendency of solvent molecule
decreases , as a result v.p. decreases. Hence according to Raoult's law
P11
P1  P1 1
o

P1
o  1
P1
P
1  1o  1  1
P1

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 o
P1  P1
o  2
P1

P1o
v.p

x
X1 = 0 X1 = 1
X2 = 1 X2 = 0

Hence relative lowering of V.P. of a solution containing non-volatile solute is equal to mole
fraction of solute in the solution.

Ideal and Non- ideal solution

 Ideal solution : The solution which obeys Raoult's law over the entire range of concentration
and temperature is called ideal solution.
 Condition/ Characteristics
 There will be no change in volume on mixing the two components ( ∆V mixture = 0)
 There will be no change in enthalpy( ∆H mixture = 0)
 Solute - solute ( A- A) , solvent - solvent ( B-B) and solute - solvent ( A - B) all
interactions are equal.
o o
 P1 = P1 X1 , P2 = P2 X 2 and PT = P1 + P2

Examples-
a) Benzene and toluene
b) n- hexane and n -heptane
c) Chlorobenzene and Bromobenzene
d) Ethyl bromide and ethyl chloride
e) Dichloro methane and chloroform
 Non-Ideal solution: The solution which does not obey Raoult's law over entire range of
concentration and temperature is called non - ideal solution.
 Condition/ Characteristics
 ∆V mixture �0
 ∆H mixture �0
 A- A, B-B �A - B
o o
 P1 �P1 X1 , P2 �P2 X 2 and PT �P1 + P2
Types of non - ideal solution
Non - ideal solutions are of two types
(a) solution showing +ve deviation from Raoult's law
(b) solution showing -ve deviation from Raoult's law

 Solution showing +ve deviation from Raoult's law

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 In this solution, solute - solvent ( A-B) interaction is less than A - A and B -B interaction.
Hence the total vapour pressure of the solution increases and solution shows positive
deviation from Raoult's law.
 Condition/ Characteristics:
 ∆V mixture > 0
 ∆H mixture > 0
 A- A, B-B > A - B
o o
 P1 > P1 X1 , P2 > P2 X 2 and PT > P1 + P2

Graphical representation ( check from ncert book)

Example-
a) Alcohol + acetone
b) Acetone + CS2
c) Acetone + Benzene
d) Methyl alcohol + water
e) Ethyl alcohol + water
f) CCl4 + CHCl3
g) CCl4 + Benzene
h) CCL4 + toluene

Explanation (Alcohol + acetone )

In pure alcohol, molecules are H- bonded. On adding acetone, its molecule get in between the
host molecule and breaks some of the H - bonds. Hence due to weakling of interaction the
solution shows positive deviation from Raoult's law.

 Solutions showing -ve deviation from Raoult's Law

In this solution, solute - solvent ( A-B) interaction is greater than A - A and B -B interaction.
Hence the total vapour pressure of the solution decreases and solution shows -ve deviation
from Raoult's law.
 Condition/ Characteristics:
 ∆V mixture < 0
 ∆H mixture < 0
 A- A, B-B < A - B
o o
 P1 < P1 X1 , P2 < P2 X 2 and PT < P1 + P2

Graphical representation ( check from ncert book)

Example-
a) Chloroform + acetone
b) Chloroform + benzene
c) Chloroform + diethyl ether

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 d) Acetone + aniline
e) HCl + H2O
f) HNO3 + H2O
g) Acetic acid + pyridine

Explanation ( Chloroform + acetone )

In acetone molecule dipole- dipole interaction is present. Similarly in chloroform molecule also
dipole- dipole interaction is present. But when they mix with each other CHCl3 molecule
is able to form H -bond with acetone molecule. Hence the escaping tendency of each
molecule decreases as a result v.p. decreases and solution shows -ve deviation from
Raoult's Law.

 AZEOTROPES

They are the binary mixture having same composition in liquid and vapour phase and boil
at a constant temperature.

It is of two types.

(a) minimum boiling azeotropes

(b) maximum boiling azeotropes

 Minimum boiling azeotropes:

 In this azeotrope A - B interaction is less than A -A and B -B interaction.

 It shows +ve deviation from Raoult's law

 The B.P. of azeotrope is less than the individual B.P.

Example :

(i) 95.37% v/v ethanol solution ( ethanol B.P. is 78.30C and H2O B.P. is 1000C and azeotrope
B.P. is 78.150C)

(ii) 71.7% propanol solution

(iii) 6.8% ethanol solution in CHCl3

 Maximum boiling azeotropes

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  In this azeotrope A - B interaction is greater than A -A and B -B interaction.

 It shows -ve deviation from Raoult's law

 The B.P. of azeotrope is greater than the individual B.P.

Example :

(i) 68% HNO3 solution (HNO3 B.P. is 359K and H2O B.P. is 373K and azeotrope B.P. is 393.5K)

(ii) 57% HI

(iii) 71.6% HClO4

Note - Azeotrope cannot be separated by distillation and fractional distillation method.

COLLIGATIVE PROPERTIES AND DETERMINATION OF MOLECULAR MASS

The properties of the solution which depend only number of solute particles present in the
solution but does not depend upon the nature of the particle are called colligative properties.

The important colligative properties are

a) Relative lowering of vapour pressure

b) Elevation in boiling point

c) Depression in freezing point

d) Osmotic pressure

 Relative lowering of vapour pressure:

When a non volatile solute is added to the solvent, the solute particle occupies in the
inter molecular space of solvent molecule as a result the escaping tendency of solvent
molecule decreases , hence v.p. decreases. According to Raoult's' law

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 P11
P1  P1 1
o

P1
o  1
P1
P
1  1o  1  1
P1

o
P1  P1
o  2
P1

o
P1  P1 n2

P1
o
n1  n2

If n2 << n1

o
P1  P1 n2
o 
P1 n1

o
P1  P1 w2 m1
o  �
P1 m2 w1

w2 m o
m2  � 1 �P1
P
1
o
 P1  w1

Question: The v.p. of pure benzene at a certain temperature is 0.850bar. A non volatile non
electrolyte solid weighing 0.5g when added to 39g of benzene, the v.p. of the solution is 0.845
bar. What is the molar mass of the solid substance?
Ans -
Note - v.p. is not a colligative property but relative lowering of v.p. is a colligative
properties.

 Elevation in boiling point:

When a non volatile solute is added to the solvent, v.p. decreases, B.P. increases. The
temperature at which the v.p. of liquid becomes equal to atmospheric pressure is called
boiling point.

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 Graph ( Check from NCERT - Book)

∆Tb α m

∆Tb = Kb x m

w2 1000
�
T2 - T1 = Kb x m2 w1

Tb
or Kb = m

 S.I. Unit of Kb is K Kg mol-1

 Where Kb is known as molal elevation constant or Ebullioscopic constant . ∆Tb

is called elevation in boiling point.

 Ebullioscopic constant can be defined as the elevation in boiling point when the
concentration of the solution is 1 molal.( when m = 1 mol Kg-1 , Kb = ∆Tb ) .

 When a non volatile solute is added to the solvent, v.p. decreases. To regain the
same v.p. more amount of heat energy is required, hence B.P. increases.

 Determination of molar mass of solute

w 1000
m2  K b � 2 �
Tb w1
Question- 18 g of glucose when dissolved in 1Kg of water, at what temperature water will boil at
1.013 bar. Kb for water is 0.52 K Kg mol-1 ?
Ans-
Question- The B.P. of benzene is 353.23 when 1.8g of a non volatile solute was dissolved in 90g of
benzene, the B.P. is raised to 354.11K. Calculate the molar mass of solute. ( Kb for benzene = 2.53 K
Kg mol-1)
Ans-

Note :
 B.P. is not a colligative property but elevation in boiling point is a colligative property.
 Same concentration of different non electrolyte have same elevation in B.P.
Ex- 6g of urea, 18g of glucose and 34.2 g of sucrose when dissolved in same amount of
solvent, have same elevation in B.P.
60g of Urea( NH2CONH2) 18g of glucose ( C6H12O6) 34.2g of sucrose( C11H22O11)
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 n = 6/60 = 0.1 n = 18/180 = 0.1 n = 34.2/342 = 0.1

 Same concentration of different electrolyte have different ∆Tb, the electrolyte having more
no. of ions have greater ∆Tb
Ex- 0.1M NaCl, 0.1M MgCl2 , 0.1M AlCl3 , 0.1M K4 [ Fe(CN)6]
2 ions 3 ions 4 ions 5 ions( greater ∆Tb)

 Depression in freezing point

When a non volatile solute is added to the solvent, v.p. decreases, B.P. increases and
freezing point decreases. The temperature at which both solid and liquid have same
vapour pressure is called freezing point.

Graph

Mathematically

∆Tf α m

∆Tf = Kf x m

w2 1000
�
T1 - T2 = Kf x m2 w1

T f

or Kf = m

 S.I. Unit of Kf is K Kg mol-1

 Where Kf is known as molal depression constant or Cryoscopic constant . ∆Tf is

called depression in freezing point.

 Cryoscopic constant can be defined as the depression in freezing point when the
concentration of the solution is 1 molal.( when m = 1 mol Kg-1 , Kf = ∆Tf ) .

 Determination of molar mass of solute

w 1000
m2  K f � 2 �
T f w1
Note :
 Frozen solvent is one type of solvent which when dissolved in solvent it does not allow to
change the temperature of the solution.

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 Ex- The running of a car in subzero weather even when the radiator is full of water has been
possible due to addition of appropriate amount of ethylene glycol in water.
 Freezing point is not a colligative property but depression in freezing point is a colligative
property.
 Same concentration of different non electrolyte have same ∆Tf value.
Ex- 6g of urea, 18g of glucose and 34.2 g of sucrose when dissolved in same amount of
solvent, have same depression in freezing point.
60g of Urea( NH2CONH2) 18g of glucose ( C6H12O6) 34.2g of sucrose( C11H22O11)
n = 6/60 = 0.1 n = 18/180 = 0.1 n = 34.2/342 = 0.1

 Same concentration of different electrolyte have different ∆Tf, the electrolyte having more
no. of ions have greater ∆Tf
Ex- 0.1M NaCl, 0.1M MgCl2 , 0.1M AlCl3 , 0.1M Urea
2 ions 3 ions 4 ( greater ∆Tf)
AlCl3 has greater ∆Tf and urea has greater freezing point.

Determination of Kb and Kf
2
R �Tb �M 1
Kb 
1000 � vap H
2
R �T f �M 1
Kf 
1000 � fus H
Where R = Gas constant
Tb = B.P. of pure solvent
Tf = Freezing point of pure solvent
M1= molar mass of the solvent
 Osmosis
 Diffusion – The mixing of particles of two matters is called diffusion.
 Osmosis- It is defined as the spontaneous flow of solvent molecules from lower
concentration to higher concentration through semi permeable membrane.
 Difference between osmosis and Diffusion

Osmosis Diffusion
Spm is used No spm is used
Only flow of solvent molecule takes place Solvent as well as solute molecule can
flow
It takes place from lower concentration to It takes place from higher concentration to
higher concentration lower concentration
It is only applicable in solution It is applicable in gaseous as well as in
solution
It can be stopped or reversed It can not be stopped or reversed

 Types of Osmosis

It is of two types

(i) Exo osmosis (solvent molecule will flow out through spm)

(ii) Endo osmosis (solvent molecule will flow inside through spm)

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 Semi permeable Membrane

It is of two types

(i) Natural spm ( vegetable membrane, animal membrane like goat membrane, dog membrane and
pigs bladder)

(ii) Artificial spm( Cellophane, parchment paper( blotting paper), Cupper ferrocyanide, silicates of
iron and cellulose acetate)

 Osmotic Pressure: It is defined as the minimum excess pressure applied on the solution side to
stop the flow of solvent molecules from solvent to solution side through spm.

Mathematically

π = CRT ( C= concentration in terms of molarity)

n
π= v RT

w2
m2 v
π= RT

dRT
m2
π=

dRT
M2 = 

Question – 200 cm3 of a solution containing 1.26g of protein. The osmotic pressure of such a
solution at 300 K is found to be 2.57 x 10-3 bar. Calculate the molar mass of the protein?

Ans-

 Advantages of Osmotic pressure over other Colligative Properties

Osmotic pressure is a best method for determining molar mass of solute due to following reasons.

 Osmotic pressure measures at room temperature where as other three Colligative properties
measure either above or below the room temperature.
 Osmotic pressure measures molarity where as other Colligative properties measures molality.
 As compare to other Colligative properties O.P. is very large for even dilute solution.
 Solutes like biomolecules, macromolecules are not stable at higher temperature.
 When the solutes are polymer they have poor solubility.

 Isotonic Solution
Two solutions having same osmotic pressure at a given temperature are called isotonic
solution.

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 Hypertonic and Hypotonic Solution
 The osmotic pressure associated with blood cell is equivalent to that of 0.9% (w/v) NaCl
solution which is called a normal saline solution.
 When a blood cell is placed in a solution containing more than 0.9%(w/v) solution, the
solvent will flow out of the cell and it shrinks. This is called hypertonic solution.
 When a blood cell is placed in a solution containing less than 0.9%(w/v) solution, the solvent
will flow in to the cell and it swells. This is called hypotonic solution.

 Reverse Osmosis ( RO) and water purification

Picture - ( Check from NCERT book)
 In this osmosis the flow of solvent molecule takes place from solution side to solvent side
through spm by application of some external pressure. This phenomenon is known as
reverse osmosis.
 RO is used for the desalination of sea water.
 Cellulose acetate is used as spm.

 Abnormal Molar mass and van’t Hoff factor

The Colligative properties of a solution depend upon the no. of solute particles present in the
solution. The various relations derived from Colligative properties are true when the solute is
non- electrolyte, But when the solute is electrolyte, the all four Colligative properties
formula are wrong due to the following two reasons.
(a) Dissociation
(b) Association
 Dissociation:
When the solute is an electrolyte and under goes ionization, the no. of particles in the solution
increases, hence Colligative properties also increases as a result molar mass decreases.
Ex- When 1 Molar NaCl dissolved in H2O , it under goes almost complete ionization and
produces two moles of ions. Hence the observed Colligative properties become twice and molar
mass of solute reduces to half ( 29.25 U) . This is called abnormal molar mass.

 Association
When the solute is an electrolyte and under goes association two or molecules associated to
form a single giant molecule. Hence the observed Colligative decrease and molar mass of solute
increases.
Ex- When 1mole of acetic acid dissolved in non polar medium benzene, it under goes
dimerisation and the molar mass of acetic acid becomes twice (120U). This is called abnormal
molar mass.
���
2CH3COOH ��� (CH3COOH)2

 van’t Hoff factor (i)

normal molar mass
i 
abnormal molar mass
abnormal c.p.(observed)
i
normal c.p. (calculated)
Pob Tb ob Tfob  normal molar mass
i    ob 
Pcal Tb cal Tfcal cal abnolmal molar mass
Hence van’t Hoff factor modified the four Colligative properties are as follows.
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 o
P1  P1
o   2 �i
P1

∆Tb = Kb x m x i

∆Tf = Kf x m x i

π = CRTx i

 When i = 1, the solute does not under goes association or dissociation ( non electrolyte)
 When i > 1 , the solute under goes dissociation
 When i < 1 , the solute under goes association
 Calculation of i
(a) For weak electrolyte-
(i) When it under goes dissociation:
CH 3COOH �� ��� 
� CH 3COO  H


1 0 0
1-α α α
i = 1- α + α + α
i=1+α
i 1
 
The general formula is n  1 , where n = no. of ions produced

(ii) When it under goes association:

���
2CH3COOH ��� (CH3COOH)2
1 0
1-α α/2
i = 1-α +α/2
i = 1- α/2
1 i
 
1
1
The general formula is n where n = no. of molecules associated

(b) For strong electrolyte

If a strong electrolyte under goes complete dissociation
i = Total no. of ions produce in a single formula unit.
Ex- NaCl , i = 2
MgCl2 , i = 3
AlCl3 , i = 4
K4 [ Fe (CN)6] , i = 5

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