SOLUTIONS
A solution is a homogeneous mixture of two of more chemically non
reacting substances whose composition can be varied within a certain limits.
In a binary solution containing only two components component present in
smaller amount is solute and the other present in larger amount is called
solvent.
Vapour pressure of a liquid /solution is the pressure exerted by the vapour in
equilibrium with the liquid /solution at a particular temperature.
Vapour pressure of liquid decreases if some non volatile solute is dissolved
in it because some molecules of solvent on the suface are replaced by the
molecules of solute.Thus the no. of solvent molecules escaping from the
surface is reduced.
Raoult’s Law
SOLUTIONS OF NON VOLATILE SOLUTES
Raoults observed that
Partial pressure of a solvent over a solution of a non volatile solute is
directly proportional to molefraction of solvent in the solution.
If A is a solvent and B is solute
XA molefraction of solvent
PA vapour pressure of solution
P0A vapour pressure of pure solvent
PA XA
PA = P0A XA
XA + X A = 1
XA = 1 -- XB
PA = P0A ( 1-- XB)
XB = P0A -- PA / P0A
PA= P0A -- P0A XB
Raoults Law:- Relative lowering of vapour pressure of a solution is equal
to mole fraction of solute.
SOLUTION OF VOLATILE SOLUTE
Partial vapour pressure or each component is proportional to mole
fraction of corresponding component.
PA XA
PB XB
PA P0A XA
PB P0B XB
P TOTAL = PA+ PB
=P0A XA + P0B XB
= P0A (1-- XB) + P0B XB
=( P0B -- P0A) XB + P0A
graph
IDEAL SOLUTION
Ideal solution is a solution
 In which each component obeys Raoults Law at all
temperatures and concentrations.
 A solution in which there is no volume change and
 And no enthalpy change on mixing.
V mix = 0
H mix = 0
( Interaction between A and B are of same magnitude as
between the pure components.)
Eg. Benzene + Toluene
Hexane + heptane
NON IDEAL SOLUTION
 Do not obey Raoults Law
 V mix ≠ 0
H mix
≠ 0
(Interaction between A and B are of different magnitude than
those in pure components.)
TYPES OF NON IDEAL SOLUTIONS
1) POSITIVE DEVIATION
 When total vapour pressure for any mole fraction is more
than that expected from Raoults Law.
 V and H are positive.
 This happens when the new interaction are weaker
than in the pure components.
A—B interaction is weaker than A—A and B—B
interaction.
Eg Ethanol and Acetone
Acetone gets in between ethanol molecules and break
H bonds of ethanol molecules.
Ethanol now will find it easier to escape than from pure
ethanol.
Graph
2) NEGATIVE DEVIATION
 When total vapour pressure for any fraction is less than that
expected from Raoults Law.
 V and H are negative.
 This happens when the new interaction are stronger
than in the pure components.
A—B interaction is stronger than A—A and B—B
interactions.
Eg Chloroform and acetone
Chloroform is able to form H bonds with acetone
molecules .
CH3
Cl
CO---------H—C----Cl
CH3
Cl
This decreases the escaping tendency of molecules of each
components.
EXAMPLES OF NON IDEAL SOLUTIONS
POSITIVE DEVIATION
NEGATIVE DEVIATION
Ethanol+Acetone
choroform + acetone
CS2 + acetone
chloroform + acetone
Benzene + Acetone
aniline + acetone
Ethanol + H2O
HCl + H2O
CCl4 + CHCl3
HNO3 + H2O
AzEOTROPIC MIXTURES
It is the mixture of two liquids which boils at a particular temperature like
pure liquids and distills over in the same composition (constant boiling
mixtures)
These are formed by non ideal solutions .
TWO TYPES
 .Azeotropic mixtures with minimum boiling point .
The boiling point of such mixture is less than either of the two pure
components.
 The boiling point of such mixtures is less than either of the two pure
components.
 This is formed by that composition of a non ideal solution showing
positive deviation for which vapour pressure is maximum.
 Eg Ethanol water mixture
Fractional distillation can concentrate it to 95% by volume of ethanol
(Azeotropic composition)Once this composition is achieved no additional
fractionation occurs.
 Boiling point of ethanol is 351.3
 Boiling point of water is 373
 Boiling point of azeotrop is 351.1
2.azeotropic mixtures with maximum boiling point
The boiling point of such a mixture is more than either of the two pure
components.
 This is formed by that composition of non ideal solution showing
negative deviation for which vapour pressure is minimum.
 Eg HNO3 + H2O (68% nitric acid and 32% water by mass.)
 Boiling point of H2O is 373 K
 Boiling point of HNO3 is 359K
 Boiling point of azeotrop is 393 K
Concentrations
Molarity =
Number of moles
Volume of solution (in cc)
x 100
Molality =
Number of moles
Mass of solution (in kg)
x 100
PPM =
Mass of A
x 106
Mass of solution
OR
= Volume of A
x 106
Volume of solution
Relation between Molarity & Molality
Molarity = n moles of solute/L
d = density of solution in kg/L
Molar mass = MB kg/moles
Mass of solute = number of moles
Mol. Mass
Therefore, Mass of solute= n x M
Mass of solution = d kg
Mass of solvent = d- n MB
Molarity =
n
d- n MB
Molarity m = 1000 x M
(1000xd)- M x GMMsolute
M = Molarity
Elevation in Boiling Point
T0 is boiling point of pure solvent. T1 is boiling point of solution
T1-T0 = ∆TB is elevation in boiling point .
Since the magnitude of elevation in boiling point is determined by lowering
of vapour pressure. Elevation in boiling point is proportional to solute
concentrations.
∆TB α ∆P, ∆P α XB
Therefore, ∆TB α XB
∆TB = K .XB
T1
P
= K
nB
nA+ nB
≈ K. nB
nA
= K. WB MA
MB WA
= nB
x MA
WA
If WA = mass of solvent in kg,
nB . = m (molality)
WA
Therefore, ∆TB = kB . m
Kb = molal elevation constant – Elevation in boiling point when one mole of
non volatile solute is dissolved per kg of solvent .
( 1 KG) Unit of kb = K kg mol-1
Calculation of Mol mass
∆TB = kb . m
∆TB = kb .x WB x 1000
WA x MB
MB = = kb .x WB x 1000
WA x ∆TB
Depression in Freezing Point
Freezing point is the temperature of which the liquid & the solid form of the
same substance are in equilibrium & hence have same vapour pressure.
Due to lower vapour pressure of solution solid form of solution separates out
at a lower temperature.
∆TF α ∆P, ∆P α XB
Therefore, ∆TF α XB
∆TF = K .XB
= K nB
≈ K. nB
nA+ nB
nA
= K. WB MA
MB WA
=
nB x MA
WA
If WA = mass of solvent in kg,
nB . = m (molality)
WA
Therefore, ∆TF = kf . m
kF = molal depression constant – molal depression constant is defined as
depression in freezing point when one mole of non volatile solute is
dissolved per kg of solvent.
MB = = kf .x WB x 1000
WA x ∆TF
Osmosis:- It is the phenomena in which there is a net flow of solvent
molecules from the solvent to the solution or from a less concentrated
solution to a more concentrated solution through a semi permeable
membrane.
Osmotic Pressure:- It is the min. pressure that must be applied on to the
solution to prevent the entry of solvent into the solution through semi
permeable membrane.
Isotonic Solution:- solution having equal molar concentration & hence equal
osmotic pressure are called isotonic solution.
0.91% solution of pure NaCl is isotonic with RBC.
Reverse Osmosis:- If the pressure applied on the solution is greater than the
osmotic pressure then solvent start passing from solution to solvent. This is
called reverse osmosis.
Determination of molecular mass
ΠV = nRT
Π = n . RT
V
= CRT
n = number of moles of solute
V = volume of solution
T = temperature
R = gas constant
C = Concentration of solution in moles/L
Π = WB RT
MB x V
MB = WB RT
ΠxV
 For macro molecule polymers & proteins, freezing point/ boiling point
are very small to be measured. Moreover heating them can change
their biological activity.
Osmotic pressure is more suitable for finding mol. mass.
Colligative Properties:- Properties of dilute solution of non volatile
solutes whose values depend upon concentration of solute particles in
solution but not on the nature of solute.
i)
Relative lowering of vapour pressure.
ii)
Elevation in boiling point
iii) Depression of freezing point
iv) Osmotic pressure
Abnormal Molecular mass:- When the molecular mass of a substance as
determined by using colligative properties does not come out to be same
as expected theoretically, it is said to show abnormal molecular mass.
Association of solute molecules :Colligative property α
1
Molecular mass of solute
.
 If solutes associates in solution number of molecules particles
decrease in the solution.
 This results in decrease in the value of colligative property.
 This leads to higher value of molecular mass than normal values.
E.g. Acetic acid
O --------- H O
C
CH3
CH3 C
O H -------- O
Acetic acid dissolved in benzene shows a mol mass of 120(normal mol
mass = 60)
Dissociation:- Electrolyte dissociates in solution to give two or more
particles. This leads to higher value of colligative property.
Since Colligative property α
1
.
Molecular mass of solute
Molecular mass calculated from colligative properties will be less than
their normal value.
E.g. NaCl dissociates to Na+ & Cl-
It is expected to give mol mass equal to half of its normal value. i.e.
58.5/2
Vont hoff factor i:In 1886 vont hoff introduced factor i
i = normal molecular mass
observed mol. mass
IF,
i = 1 - No association or dissociation takes place.
i < 1 - Association takes place.
i > 1 - Dissociation takes place.
∆TB = i kB . m
∆TF = i kF . m
ΠV = i nRT
P0A –PA = i XB
P0A
Since colligaitive property are inversely proportional to mol. mass.
i = normal colligative property
.
observed colligative property
i=
number of particles after association or dissociation
number of particles before association or dissociation
Degree of dissociation:- α – The fraction of total molecules which
dissociates into ions.
Expression for dissociation isα = i-1
n-1
n = no. of particles formed on dissociation
number of moles dissociates = α
number of moles left undissociates = 1- α
total number of dissociated particles = n α
total number of partcles at eqlm = (1- α) + n α
= 1+ (n-1) α
i = no. of particles after dissociation .
no. of particles before dissociation
i = i + (n-1) α
1
α = i-1
n-1
NaCl → Na+ + Cl(1- α) α
α
(1- α, 2 α) at eqlm.
Expression for associationα = (i-1)
1 -1
n
after association
nA → An
1-α
α
n
number of association moles = α
number of unassociated moles = 1- α
total number of moles = 1- α + α
n
i= no. of particles after association .
no. of particles before association
i = 1- α + α
n
1
= 1+ α ( 1 – 1)
n
Therefore, α = (i-1)
1 -1
n
OR
α = (i-1)n
1-n
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