Prepared By Dr.
Ramesh Kumar Behera
Solution = Solute + Solvent ( Large quantity ) Solvent decide the physical state of solution .
1 ppm of fluoride ions in H2O prevents tooth decay. while 1.5 ppm causes the tooth to become
mottled and high concentration of flouride ion can be poisonous. NaF is used as rat poison
Types of solution
Depending on the state of solution, the solution can be classified as- Gaseous, Liquid and solid
Solution Solute Solvent Examples
Gaseous Solid Gas Camphor in nitrogen gas
Liquid Gas Chloroform mixed with nitrogen gas
Gas Gas Mixture of oxygen and nitrogen gas
Liquid Solid Liquid Glucose dissolved in water
Liquid Liquid Ethanol dissolved in water
Gas Liquid Oxygen dissolved in water
Solid Solid Solid Copper dissolved in gold
Liquid Solid Amalgam of mercury with sodium
Gas Solid Solution of hydrogen in palladium
Concentration of solution
Mass of component in solution Commercial bleaching solution
Mass percentage ( W/ W ) = X 100 contains 3.62 mass percentage of
Mass of solution sodium hypochlorite in water
Volume of component in solution 35% (v/v) solution of ethylene glycol,
Volume percentage ( V / V ) = X 100 an antifreeze is used in cars for
Volume of solution cooling engine.At this concentration
the antifreeze lowers the freezing
point of water to 255.4 K ( -17.60 C )
Mass of component in solution X 100
Mass by volume percentage =
Volume of solution
Sea water contains 5.8 ppm dissolved
No. of parts of component oxygen i.e. 5.8 gm of oxygen present
Parts per million ( PPM ) = X 106
Total no. of parts of solution in 106 gm of sea water.
Suppose one solution contains
No.of moles of the component n1 moles of solvent
Mole fraction ( X ) =
Total no. of moles of the solution n2 moles of solute
Calculate the mole fraction of ethylene glycol of 20 % by mass of solution n1
Then M.F of solvent ( X1 ) =
n1 + n2
Calculate the mole fraction of benzene in solution containing 30% by
n2
mass of carbon tetrachloride. Then M.F of solute ( X2 ) = n
Calculate the molarity of a solution containing 5 gm of NaOH in 450 ml
1 + n2
X1 + X2 = 1
of solution.
No. of moles of solute
Molality ( m ) =
Weight of solvent in kg
2.5 gm of acetic acid present in 75 gm of benzene, calculate molality of the solution .
Calculate molality, molarity and mole fraction of KI, if the density of 20% ( mass/mass ) aqueous KI is 1.202 g/ml
Page - 1
rkbchem3@gmail.com Page 1
Solubility It represent the maximum amount of a substance that can be dissolved in a specified amount of
solvent at a particular temperature.
S olubility depends upon:-
[1] Nature of solute and solvent Like dissolves like
i.e. Polar solute dissolves in polar solvent e.g.- NaCl dissolves in water.
and non- polar solute dissolves in non-polar solvent e.g.- wax dissolves in kerosene.
[2] Temperature
For S olute +S olvent S olution , ∆sH = - ve
Increase of temperature decreases the solubility.
For S olute +S olvent S olution , ∆sH = + ve
Increase of temperature increases the solubility.
[3] Pressure Pressure does not have any significant effect on solubility
of solids in liquids , because solids and liquids are highly
incompressible and remain unaffected by change in
Henry's law pressure.
This law states that at constant temperature, the solubility of a gas in a liquid is directly proportional
to the pressure of the gas. OR
The mole fraction of gas in the solution is proportional to the partial pressure of the gas over the solution.
OR
The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction (X) of the gas in the solution
P α X P = KH X
∴ KH Henry's constant
The value of KH is a function of nature and temperature of the gas.
Higher the value of KH at a given pressure, lower is the solubility of P S lope = KH
the gas in the liquid.
W ith increase of temperature, solubility of gas decreases i.e. K H value increases.
X
KH value of gas 'A' is 69.16 kbar, gas 'B' having 88.84kbar. Between A & B which one is solouble in water with more extent ?
G as 'A' will be more soluble,because it has lower KH value
If nitrogen gas is bubbled through water at 298 K,how many milli moles of nitrogen gas would dissolve in one
litre of water ? Partial pressure of nitrogen is 0.987 bar,Henry's constant for nitrogen at 293K is 76.48 kbar.
Raoult's law
For a solution of volatile liquid, the partial pressure of each component in the solution is directly
proportional to its mole fraction.
Partial pressure α Mole fraction of volatile component Solvent P0 P
Vapour pressure
Let a solution contains, A Volatile solvent ( M.F = X 1 )
B Non-volatile solute ( M.F = X 2 )
Vapour pressure of solution α Mole fraction of solvent. ∆P Solution
P α X1 P = P 0X P0
∴V.P of pure solvent
1
Temperature
P = P 0(1 - X 2 ) P = P 0 - P 0X 2
P 0X 2 = P 0 - P
P0 -P ∆P
X2 = X2 =
P0 P0
∴ ∆P = P 0 - P ∆P
Lowering of vapour pressure = R elative lowering of vapour pressure.
P0
Raoult's law also states that relative lowering of vapour pressure is equal to mole fraction of solute.
Page -2
rkbchem3@gmail.com Page 2
S pecial case - If both solute and solvent are volatile in nature
A Volatile solvent ( M.F = X 1 ) having v.p = P 10
B Volatile solute ( M.F = X 2 ) having v.p = P 20
P o o
Let in the solution of A &B , the partial pressure of 'A' is P 1 S lope = P2 P1
o
and partial pressure of 'B' is P 2 P 1 = X1 P 1 P2 = X2 P2o o
P1
o
V.P of solution P
= P1 + P2 +
o
= X1P1 X2 P2
X2
o o o o
= P1 ( 1 - X2 ) + X2 P2 = P1
o
P1 X2 + X2 P2
o o o
i.e. P = ( P2 P1 ) X2
+ P1 straight line equation y = mx + c
Composition of mixture in liquid and vapour phase
P1
Mole fraction of 'A' Y1 =
P Gaseous Phase
Mole fraction of 'B' P ( Dalton's law of partial pressure )
Y2 = 2
P
Total pressure P = P1 + P2
Pure Component A B Liquid Phase
o o ( Raoult's law )
V.P of Pure Component P1 P2
Mole Fraction X1 X2
o o
Partial pressure P1 = X1 P1 P2 = X2 P2
The vapour pressure of pure liquid A & B are 450 and 700 mm Hg respectively at 350 K. Find out the composition of the
liquid mixture, if total vapour pressure is 600mm Hg. Also find the composition in vapour phase.
Ideal solution
[1] The solutiopn which obeys R aoult's law under all conditions of temperature, pressure and
concentration is known as ideal solution.
[2] In ideal solution enthalpy of mixing and volume of mixing is zero i.e. ∆H mix = 0 ∆Vmix = 0
[3] In this case the intermolecular attractive forces between solvent-solute molecules are same with
solute-solute and solvent-solvent molecules
Practically, there is no solution which behaves strictly as an ideal solution.
Examples, solutions of :- [1] n-hexane & n-heptane
[2] Chlorobenzene & Bromobenzene
[3] Ethyl bromide & Ethyl iodide
[4] Carbon tetra chloride & silicon tetra chloride
[5] Benzene & Toluene
The plot of vapour pressure and mole fraction of an ideal solution ( Benzene-Toluene ) at constant
temperature 0
0
P1
Let v.p of pure benzene = P1 0 P= P
and v.p of pure toluene = P2 1 + P
2
In the solution M.F of benzene X1
P1 P20
M.F of toluene X2
0 V.P
P20
Bezene is more volatile than toluene so, P1
> P2
The vapour pressure of solution increases with increase of mole
fraction of more volatile component
X1 = 1 X1 = 0
X2 = 0 M.F X2 = 1
Benzene Toluene
Non-ideal solution
The solutiopn which does not obey R aoult's law under all conditions of temperature, pressure and
concentration is known as non-ideal solution.
i.e. ∆H mix = /0 ∆Vmix = 0 / Page -3
rkbchem3@gmail.com Page 3
P= P
Non-ideal solution showing positive deviation 1 + P
In this case the intermolecular attractive forces between 0
P1 2
solvent-solute molecules are weaker than those between
solute-solute and solvent-solvent molecules
i.e. ∆H mix >0 and ∆Vmix 0> P1 P20
solutons of :- H2O + Ethanol H2O + Cyclohexanal V.P P2
C6H6 + CCl 4 H2O + Cyclohexane
CS2 + Acetone Acetone + Ethanol
X1 = 1 X1 = 0
CHCl 3 + C2H5OH CCl 4 + CHCl 3 M.F
X2 = 0 X2 = 1
shows positive deviation
e.g. Incase of H2O + Cyclohexane ,the H-bonding present in water molecules are cut off by adding
cyclohexane,hence the vapour pressure of the solution increases.
Non-ideal solution showing negative deviation
In this case the intermolecular attractive forces between 0
P1
solvent-solute molecules are stronger than those between
solute-solute and solvent-solvent molecules
P= P
i.e. ∆H mix
< 0 and ∆Vmix 0 < + 1 P2
P20
P1
solutons of :- CH3COOH + Pyridine CHCl 3 + Acetone V.P
H2O
+ H2SO 4 H2O
+ HCl
P2
H2O
+ HNO3 Acetone + Aniline
X1 = 1 X1 = 0
X2 = 0 M.F X2 = 1
e.g. Incase of CHCl 3 + Acetone ,new intermolecular attraction arises between and CHCl 3 Acetone
due to formation of hydrogen bond.
Azeotropes
Azeotropes are defined as the mixture of liquids which boil at constant temperature like a pure
liquid and possess the same composition of the components in the liquid as well as in vapour phase.
Azeotropes are mixtures and not compounds,because both the boiling point and composition of azeotropes
is changed whereas for a chemical compound, composition remains constant over a range of pressure .
There are two types of Azeotropes :-
[1] Minimum boiling point azeotropes
The solution which shows positive deviation, form minimum boiling point azeotropes at a specific
composition. e.g. 95% by volume of ethanol with water boil at 351.15 K
[1] Maximum boiling point azeotropes
The solution which shows negative deviation, form maximum boiling point azeotropes at a
specific composition. e.g. 68% by mass of nitric acid with water boil at 393.5K
Colligative Properties
The properties of solution that depends upon the number of solute particles irrespective of their
nature is known as colligative properties. ∆P
[1] R elative lowering of vapour pressure = X2 [2] Elevation in boiling point ( ∆Tb )
P0
[3] Depression in freezing point ( ∆Tf ) [4] O smotic pressure (π )
Elevation in boiling point ( ∆Tb ) 1 atm
O
Elevation in boiling point ( ∆Tb ) = Tb Tb
Vapour pressure
Solvent
O
Tb Boiling pont of solvent Tb Boiling pont of solution
Solution
∆Tb α ∆P W e know ∆P α m ∆Tb
O
Tb Tb
S o ∆Tb α m S o ∆Tb = Kb m Temperature
W here Kb is proportionality constant , known as Ebullioscopic constant / boiling point elevation
constant / molal elevation constant.
Page -4
rkbchem3@gmail.com Page 4
∆Tb
Kb = If 1 mole of solute present in 1 kg of solvent [ molality ( m =1) ] Kb = m
m
Molal elevation constant ( K ) bis defined as the elevation of boiling point of one molal solution.
Kb value depends upon the nature of solvent, For water =K
0.52
b K.kg.mol-1
18 gm glucose dissolved in 700 gm of water, at what temperature the solution will boil at 1.013 bar pressure.
Vapour pressure of pure benzene at certain temperature is 0.850 bar. A non-volatile and non-electrolyte solid
weighing 0.5 gm when added to 39gm of benzene, vapour pressure changes to 0.845 bar. What is the molar
mass of the solute?
Depression of freezing point ( ∆Tf ) Solvent
0 Solution
Depression of freezing point ( ∆Tf ) = Tf - Tf Liquid
0
Solid
Vapour pressure
W here Tf Freezing point of solvent Tf Freezing point of solution
Freezing point is the temperature at which V.P of solid = V.P of liquid of same substance.
∆Tf
∆Tf α ∆P W e know ∆P α m
0
S o ∆Tf α m S o ∆Tf = Kf m Tf Tf
Tem perature
W here Kf is proportionality constant , known as cryoscopic constant / freezing poin depression
constant / molal depression constant. ∆Tf
Kf =
m
If 1 mole of solute present in 1 kg of solvent [ molality ( m =1) ] Kf = ∆Tf
Molal depression constant ( K ) is
f defined as the depression of freezing point of one molal solution.
Kf value depends upon the nature of solvent, For water Kf K.kg.mol-1
= 1.86
45 gm of ethylene glycol is mixed with 160 gm of water. Calculate the freezing point of solution.
Freezing point of an aqueous solution is - 50 C , Calculate its boiling point.
Osmosis
The spontaneous flow of solvent molecules from low concentrated side to high concentrated side
through a semipermeable membrane is known as osmosis.
[ In diffusion, in order to maintain equal concentration the solute particles flow from high
concentrated side to low concentrated side.]
V.P of low concentrated side > High concentrated side Flow of solvent π Piston
molecules
S o S olvent molecules flow from low concentrated side
to high concentrated side
Osmotic pressure ( π ) Low concentration High concentration
The pressure exerted at high concentrated side due to Hypotonic soln Hypertonic soln
osmosis is known as osmotic pressure.
[1] If two solutions having same concentration
i.e. same osmotic pressure is known as Isotonic solution S em iperm eable
[2] The low concentrated solution is known as Hypotonic solution m em brane
[3] The solution having high concentration i.e. higher osmotic
pressure is known as Hypertonic solution.
O smotic pressure is directly proportional to the concentration of the solution i.e. π α C
O smotic pressure increases with increase in temperature.i.e. π α Τ
S o π α CT π =CR T
200 ml of an aqueous solution of a protein contain 1.26 gm of protein. The osmotic pressure of such a
solution at 300K is found to be 2.57 Χ 10-3 bar. Calculate the molar mass of the protein.
Page -5
rkbchem3@gmail.com Page 5
Edema
W hen we take a lot of salty food , water retension in tissue cells and intercellular spaces
increases because of osmosis. The resulting puffiness or swelling is called edema.
Reverse osmosis
The phenomenon of movement of solvent molecules from high concentrated side to low concentrated
side through semipermiable membrane by applying pressure on the high concentrated side which is
greater than osmotic pressure is known as reverse osmosis.
R everse osmosis is used in desalination of sea water.
Van't Hoff factor
Calculated colligative properties and observed colligative properties some times differ due to
abnormal molecular masses.
[1] The abnormal molecular masses results due to molecular dissociation or molecular association.
[2] In order to equalise the observed and calculated colligative properties, we should multiply
Van't Hoff factor (i) with the calculated properties.
Observed colligative
Calculated colligative
∆P = i X 2 P 0 O bserved colligative properties
∆Tb = i Kbm
i = Calculated colligative properties
properties
∆Τf = i Kf m
properties
i =
No. of particles after association / dissociation
π = i CR T
No. of particles before association / dissociation
W
Colligative properties α no.of moles , No. moles =
M
1
S o colligative properties α Calculated molecular mass
M i = O bserved molecular mass
Molecules C6H12O 6 NaCl CH3COONa 2CH3COOH Na2SO 4 K4[Fe(CN)6]
Van't Hoff Factor ( i ) 1 2
3 5
1 2
2
Relationship between Van't Hoff factor ( i ) and degree of dissociation ( α )
AB x y x Ay+ + y B x- No. of particles
1 0 0 Before dissociation = 1
1-α xα yα After dissociation = 1- α + xα + yα
= 1- α + α ( x +y )
1- α + α n = 1- α + α n (n- Total no. of ions)
Van't Hoff Factor ( i ) = 1
1−i
i = 1- α + α n i = 1- α ( 1 − n ) α(1−n)= 1−i α= 1−n
Relationship between Van't Hoff factor ( i ) and degree of association ( β )
nA An No. of particles
1 0 Before association = 1
β 1
1−β β
After association = 1−β + n =1−β 1− = 1 −β n −1
n n n
1−β n−1
n n−1 n−1
Van't Hoff Factor ( i ) = 1
i =1−β n
1 − i = β
n
β = n ( 1 -i )
0.6 ml of acetic acid having density 1.06 gm/ml is dissolved in 1L of water. If Van't Hoff
n −1
factor is 1.041, Calculate dissociation constant of acetic acid.
2 gm of benzoic acid dissolved in 25 gm of benzene shows a depression in freezing point equal to 1.62 K. Molal
depression constant for benzene is 4.9 Kkg/mol.W hat is the percentage of association of acid if it forms dimer in
solution.
Which colligative property is used for the determination of molar masses of macromolecules?
O smotic pressure is used for the determination of molar masses of macromolecules.Because,
[1] It is performed in room temperature with help of concentration expressed in molarity.
[2] Compared to other colligative properties, its magnitude is large even for very dilute solution.
[3] Macro molecules are unstable and poor solubility at higher temperature.
Page -6
rkbchem3@gmail.com Page 6
