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PHYSICAL CHEMISTRY – Two Components Liquid System

IMMISCIBLE LIQUIDS
Immiscible liquids are liquids which do not mix up to form homogeneous mixture.
When there are two immiscible liquids they form a so called immiscible pair.
Immiscible liquids form heterogeneous mixture.
For the immiscible liquids, the intermolecular force of attraction is greater compared to intermolecular forces of attraction , that’s why the liquids don’t mix up.
Since the liquids do not mix up, than their total vapor pressure (PT) is equal to the sum of the pure vapor pressure of the components.
In immiscible liquids normally the denser component is found at the bottom (lower layer) while the less denser component is floating on the denser component (upper layer).
Example
Consider the immiscible pair of components A and B in which A denser than B
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
The immiscible pair is kept in a separating funnel in order to see them clearly.
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Partition law states that
“When a solute that is soluble in both liquids is added, then it will dissolve”
The ration of concentration of solute added in a pair of a immiscible liquid is constant
i.e
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
The distribution of solute in a pair of immiscible liquid is governed by the Partition law or Distribution law.

PARTITION LAW
Partition law state that;
“In a pair of immiscible liquid, when solute which is soluble in both is added, it will distribute itself in such a way that the ratio of its concentration between the two liquid is constant.”
The constant in distribution law is termed as Distribution coefficient, Distribution constant or Partition constant. It is denoted by Kd or KD.
Note
The solute added in a pair of immiscible mixture can be in either of three states (liquid, gases, solid).

APPLICATION OF PARTITION LAW
One of the application is the extraction of solute from one component by mixing the solution with the second liquid that has no solute at all.
The liquid component which is removing the solute is called Extracting component and that in which the solute is removed from is called Extracted component.
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
In terms of Extractions
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Note
Concentration is the amount of substance per unit volume.
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Also
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
NOTE
During extraction of solute the amount of solute in extracted component will be decreasing while the amount of extracting component will be increasing.

Example 1
a) State the partition law
Partition law state that;
“When a soluble solute is added in a pair of immiscible liquids, it will distribute itself in such a way that the ratio of its distribution between the two liquids is constant.”
b) What does the terms “Partition coefficient” mean?
Partition coefficient is the concentration of solute in immiscible liquids.
c) An aqueous solution containing 10g per litre of solute X. This solution was shake with 100cc of ether, on shaking 6g of X was extracted. Calculate the amount of X extracted from aqueous so residue after shaking with 100cc of ether.
Solution
Given
Extracted component = 10g/ l
Volume of extracting component = 100cc
Mass of X in water (aqueous solution) = 10g
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Kd = 15
Let the amount extracted be ‘a’
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
60 – 15a = 10a
60 = 10a + 15a
60 = 25a
a = 60/25
a = 2. 4g/cc
The amount of X extracted from the aqueous residue is 2. 4g/cc.
NOTE
If layers are not specified then the word “between” shows the numerator and denominator of the formula.

Example 2
A solid X is added to a mixture of benzene and water after shaking well and allowing to hand, 10ml of benzene layer was found to certain 0. 13g of X and 100ml of water layer contained 0.22g of X.
Calculate volume of distribution coefficient of solute X between benzene and water layer
Solution
Mass of solute X in benzene = 0.13g
Volume of benzene = 10ml
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)

Example3
In the distribution of succinic acid between ether and water at 15oC , 20ml of the ethereal layer contains 0.092g of the acid. Find out the weight of the acid present in 50ml of the aqueous solution in equilibrium with it. If the coefficient kb between water and ether is (1.196g) and kd is 5.2
Solution
Mass of succinic acid in ethereal = 0.092g
Volume of ethereal = 20ml
Conc. Of succinic acid in ethereal = 0.092/20
= 4. 6 X 10-3 g/mol
Coefficient of succinic acid between water and ethereal
= 5.2
Volume of water = 50ml
Let X be the weight of succinic acid in water
Conc. of succinic in water = W/50
From;
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
The weight of the acid represent in aqueous solution 1.196g
Example 4
An aqueous solution of succinic acid at 15oC containing 0.07g in 10ml is in equilibrium with an ethereal solution which has 0.013g in mo. The acid has its normal molecular weight in both solvents. What is the concentration of the ethereal solution which is in equilibrium with aqueous solution containing 0.024g in 1oml?
(Ans: 0.00044g/ml)
Solution
Mass of succinic acid in aqueous solution = 0.07g
Volume of aqueous solution = 10mls
Concentration of succinic in ethereal = 0.013
Concentration of succinic in solution = 0.07/10
= 7 X 10-3
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Kd = 5.38
Now
For second extraction
Mass of Aqueous solution = 0.024g
Volume of Aqueous solution = 10ml
Conc. Of succinic in aqueous solution = 0.024/10
= 2.4 X 10-3
Volume of ethereal = 10ml
Let X be the concentration of succinic acid in ethereal
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
For more than one extraction we normally use the following formula;
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Where by;
Are is the amount of solute remain
Va is the volume of extracted solution
Vb is the volume of extracting solution
K is the constant of distribution
Wo is the original weight of the solute
From the formula, the amount extracted can be calculated as ( A ex)
Aex = Wo EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)Are
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
If the amount extracted and amount remained are known, then their respective percentages can be calculated.
i.e
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
For extracted %
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Example 1
Solute the partition law
Calculate the amount of solute extracted by shaking 1 litre of aqueous solution containing 11g of Q with;
i) 100ml of ether ( 10g)
ii) Two successive volume of 50ml of ether (kd = 100 ) ( 10. 69g )
Answer
The partition law states that;
“When a soluble solute is added in a pair of immiscible liquid, it will distribute itself in such a way that, the ration of its distribution in the liquid is constant”
Data given
Mass of solute Q = 11g
Volume of aqueous solution = 1 litre = 1000cc
Conc. Of Q in aqueous solution
Let X be the amount extracted
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
10x = 1100 – 100x
110x = 1100
x = 10g.
The solute extracted is 10g.
b) (ii) Number of extraction = 2
volume of ether ( Vb ) = 50ml.
volume of water ( Vb ) = 1000ml.
kd = 100
Wo = 11g
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
= 0.30556
Again
Aex = Wo – Are
= 11 – 0.3556
= 10.69g
The solute extracted is 10.69g.

Example 2
a) Explain the principle of solvent extractions.
b) What is the condition necessary for solvent extraction?
Answer
b) The conditions of solvent extraction;
(i) The liquid to be mix must form immiscible solution.
(ii) the solute that is added to the extracted component must be soluble to the extracting component like wise.
a) The principles of solvent extraction;
i) Division of the volume of extracting components
So as the extraction to be efficient the extracting components can be divided into two or more partitions.
ii) When a liquid A ( extracting component ) mixed with liquid B ( extracted component) must form the immiscible liquid with layer between them.
iii) The solute should be soluble in both liquid component hence it will allow its distribution such pair of immiscible liquid.

Example 3
a) State the partition law
Two form five girls each were given a solution which contains 10g of solute A in 900cc of solvent C. The first girl used 900cc of solvent B to extract solute from C. The second girl decided to use 300cc of B for the three extractions.
The distribution coefficient of solute A between C and B is 8.
i) Calculate the percentage of A left in C by the first girl.
ii) Calculate the percentage of A left in C by the second girl.
iii) Comment on the result obtained by the two girls.

Solution
a) Partition law states that:
“When a soluble solute is added in a pair immiscible liquid it will distribute itself in such a way that, the ratio of concentration of it in each component is constant”.
b) Solution
Mass of solute Wo = 10g
Volume of extracted = 900c
c
Volume of extracting Vb = 900cc
Distribution constant Kd = 8
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
= 1.11g
% remained = 1.11/10 X 100 = 11%
ii) % remained by the second girl
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
= 2%
The percentage of A left inn C by the second girl is 2%.
iii) For a successful extraction, the extracting liquid component must be divided into small portion as possible.

Example 4
Find out the principles of solvent extraction 100cm3 of there is available for extracting the solute X from 100cm3 of water. The partition coefficient of X between ether and water is 4.
i) Calculate the fraction of X extracted by using 100cm3 of ether all at once.
ii) Calculate the fraction of X extracted by using 4 ( If aqueous solution contains 8g of X )

Answer
The principles of solvent extraction.
i) The solute that is added in the immiscible mixture must be soluble to both liquids
So as the extraction to be perfect the extracting liquid must be divided into small portions as possible.
ii) When the extracted liquid is mixed with the extracting liquid, they should form immiscible mixture.
Solution
i) X extracted
given:
volume of extracting Vb = 100cm3
volume of extracted Va = 100cm3
Distribution constant k = 4
mass of solute X = 8g
from
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
The fraction of X extraction is
ii) Fraction of X extracted by using two 50cm3
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
= 8/9
Amount extracted = 8 – 8/9
= 64/9 g
The amount of X extracted by using 2, 50cm3 is 64/9 g.

Example 5.
The Mogul oil company is disturbed by the presence of impurity M In its for star petrol. 1 litre of petrol contains 5g of M. In an effort to decrease the concentration of M in a petrol, Mogul has discovered the secret of solvent S and the partition coefficient of M between petrol and S is 0.01.
a) What is meant by the term partition?
b) Explain the principle of solvent extraction
c) Calculate the total mass m of M removed in by using
i) One portion
ii) Two 50cm3 portion of solvent

Solution
a) Partition is the distribute of solvent in a pair of immiscible mixture.
b) i. The principle of solvent extraction;
“When the two components are added (extracted and extracting) they should form immiscible mixture.”
ii. The solute that is used must be soluble in both extracting and extracted component.
iii. The extracting component should be divided in small portions so as the experiment to be perfect.
c) Given
Volume of petrol Va = 1 litre = 1000cm3
Mass of M X = 5g
Distribution constant kd = 0.05
Volume of solvent S Vb = 100cc
i)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
The mass removed is 4.55g
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
= 0.138g
The amount removed is 5 – 0.138g = 4.86g.
SEPARATION OF IMMISCIBLE LIQUID
Immiscible liquids are separated by the process of steam distillation.
Steam distillation.
This is the process of separating immiscible liquids of different boiling points by passing super heated steam through
Condition necessary for steam distillation
In order for steam distillation to be feasible, the following conditions are;
i) The two liquids should have different B.P.
ii) The two liquids should be immiscible.
iii) There should be no volume change.
iv) The total vapor pressure of the liquids should be equal to the sum of the components vapor pressure.

APPLICATION OF STEAM DISTILLATION
Steam distillation can be used to determine the molar mass of unknown liquid.
Let the two liquids A and B form an immiscible pair
And
nA = a
nB = b
Since the two liquids are immiscible then, their distillation process can be explained in terms of their proportions or compositions (mole fraction).
From the number of moles of components the total number of moles can be obtained
i.e nT = nA + nB
nT = a +b
If nT is known then mole fraction or composition can be calculated
i.e
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
For immiscible liquids , the ratio of their compositions is equal to the ratio of their vapor pressure.
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Where:
MB is the molar mass of B
MA is the molar mass of A
mA is the mass of A
mB is the mass of B
PA is the vapor pressure of A
PB is the vapor pressure of B

Example 1.
A solution of 6gm of substance X in 50cm3 of aqueous solution is in equilibrium at room temperature with an ether solution of X containing 108gm of X in 100cm3 . Calculate what weight of X could be extracted by shaking 100cm3 of an aqueous liquids containing 10gm of X with;
i) 100cm3 ether
ii) 50cm3 of ether twice at room temperature.

Solution
Concentration of X in H2O = 6 g/50 cm3
Concentration of X in ethereal = 108/100
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
K = 9
Now
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)

Example 2.
What is steam distillation?
State the four conditions necessary for steam distillation.
A organic liquid distills in steam, the partial pressure of the two liquids at the boiling point are 5.3 k pa for organic liquid and 96 k pa for water. The distillate contains the liquids in the ratio of 0.48g organic liquid to 1g of water. Calculate the molar mass of organic liquid.

Answer
a) Steam distillation Is the process of separating immiscible liquids of different boiling points by passing super heated steam through it.
b) The conditions necessary for steams distillation are;
i) The liquids must form immiscible solution.
ii) The total vapor pressure is equal to the sum of the vapor pressure of the components.
iii) There should be no change in volume.
iv) The liquids should have different boiling point.

Solution
Po = 5.3kpa
mo = 0.48
Mo = ?
Pw = 96 kpa
Mw = 18
mw = 1g
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Note
Unit conversion
i) 1 N/m2 = 1 pa
ii) 1 atm = 760 mmHg
iii) 1 atm = 1.01 EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)105 Pa
iv) 1 atm = 1.01 EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)105 N/M2

Example 3.
a) Differentiate between thermal distillation and steam distillation.
b) Bronbenzene (C6H5Br) distills in steam at 95oC the vapor of Bromobenzene and water are 1.39 x 104 N/M2 and 8.5 X 104 N/M2 .Calculate the percentage by mass of bromobenzene.
(c =12 Br =80 O = 16 H = 1) Note mass of H2O =24g
Solution
a) Thermal distillation is the process of separating immiscible mixture by the use of thermal energy (heat) while
Steam distillation is the process of separating immiscible liquids by having different boiling points by passing super heated steam through it.
Data
PB = 1.39 X 104 N/M2
Pw = 8.5 EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)104 N/M2
MB = 157
Mw = 18
mw = 24g
from
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)

Example 4.
An organic liquid Q which do not mix with water distills in steam at 96oC under the pressure of 1.01 EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1). The pressure of water at 96oC is 8.77 EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)N/M-2 . The distillate contains 51% by mass Q . Calculate the molar mass of Q.
Solution
Atmospheric pressure Patm = 1.01 EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)N/M2
PQ = Patm EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)Pw
Pw = 8.77 EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Now
PQ = 1.01 EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)– 8.77 X 104
= 1033EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
MQ = ?
Mw = 18
Mass of Q mB = 51g
Mass of W mw = 49
From
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Example 5.
i) Define the term steam distillation.
ii) State the conditions necessary for steam distillation.
Calculate the molar mass of the compound B whose mixture with water distills at 95oC. At this temperature the pressure of compound B and water are 119mmHg and 64mmHg. The ratio of B to water is 1.61 : 1

Answer
i) Steam distillation is the process of separating immiscible mixture by passing super heated steam through it.
ii) The condition necessary for steam distillation are
There should be no change in volume
The liquid must form immiscible solution
The liquids should have different boiling point
The total vapor pressure is equal to the sum of pressure in the mixture.

Solution
PB = 119mmHg
PW = 64 mmHg
MB = 1.61g
mw = 1g
MB = ?
Mw = 18g/mol
From
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Example 6.
At a pressure of 760 mmHg, a mixture of nitrobenzene (C6H5NO2) and water boils at 99oC. The vapor pressure at this temperature is 733 mm. Find the proportion of water and nitrobenzene in the distillate obtained by steam distillation of pure C6H5NO2.

Solution
Atmospheric pressure patm = 760
Pressure of H2O Pw = 733
Pressure of nitrobenzene pn = Patm – Pw
= 760EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1) 733
Pn = 27
Mw = 18
Mn = ( 12 EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)6) + 5+ 14 ( 16EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1) 2)
Mn = 123
From
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Example 7.
A mixture of water and bromobenzene (C6H5Br) distills at 95oC and the distillate contain 1.6 times as much C6H5Br as water by mass. At 95OC the vapor pressure of water and C6H5Br are 640mmHg and 120mmHg respectively. Calculate the molecular weight of bromobenzene.

Solution
PW = 640mmHg
PB = 120mmHg
Mw = 18g/mol
MB = ?
Let X be the mass of water (Mw).
1.6X will be the mass of C6H5Br (MB).
From
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)

COLLIGATIVE PROPERTIES
What is colligative properties?

Definition
Colligative properties are properties of the liquid which change depending on the number of particle of solute added, but not on the nature of the solute.
Mainly there are 4 colligative properties:
i) Lowering of vapor pressure.
ii) Boiling point elevation.
iii) Freezing point elevation.
iv) Osmotic pressure.
Assumptions of colligative properties:
i) The solute should not react with solvent
ii) The solute should be not volatile compared to solvent
iii) The solute should not dissociate or associate in the solvent
1. LOWERING OF VAPOR PRESSURE
Vapor pressure is the pressure exerted by vapor against the atmospheric pressure.
Lowering of vapor pressure: Is the difference between the original pressure of liquids solvent (Po) and the pressure of the solution.
Effect of solute on the vapor pressure of the solvent
When solute particles are added in the solvent, the vapor pressure of the solution is lowered.
When solute particles dissolve in a given solvent normally they collide with the solvent molecule and hence prevent / decrease the number of solvent molecules that escape from liquid phase to vapor phase. This causes the decrease amount of vapor above the solution and normally causes the decrease in vapor pressure.
Relative lowering of vapor pressure
Relative lowering of vapor pressure is the ratio lowering vapor pressure to the vapor pressure the solvent.

RAOULT’S LAW OF VAPOR PRESSURE
It states that “The relative lowering of vapor pressure is proportional to the mole fraction of the solute added”

Mathematically
Let P o be the vapor pressure of the solvent
P be the vapor pressure of solution
Xs be the mole fraction of solute
From Raoult’s law of vapor pressure
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
For mole fraction of solute
Let n be number of mole solute
N be number of mole solvent
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
The molar mass of solute can be determined
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Example 1.
What do you understand by the term “ Colligative property”?
Colligative property is the property of a liquid which change depending on the number of properties of solute added but not on the nature of the solute.
A solution is prepared from 90g of water and 10.6g of non – volatile solute. If the vapor pressure of the solution at 60oC was found to be 0.1867 atm. Calculate the relative molecular mass of solute. Given that V. P of H2O at 60oC was 0.1966 atm
Solution
Mass of water (solvent) = 90g.
Mass of non – volatile solute = 10.6g.
Vapor pressure of solution (P) = 0.1867 atm.
Vapor pressure of solvent Po = 0.1966 atm.
Recall
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Example 2.
When 114g of sucrose are dissolved in 1000g at water the vapor pressure was lowered 0.11mmHg . Calculate the relative molecular mass of sucrose if the vapor pressure of water at 20oC was 17.54 mmHg.
Solution
Mass of solute = 114g
Mass of solvent = 1000g
Lower vapor pressure = 0.11mmHg (Po – P)
Vapor pressure of solvent = 17.54mmHg
Recall

EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)

Example 3.
Calculate the vapor pressure lowering caused by the addition of 100g of sucrose of molar mass 342g/mol to 1000g of water if the vapor pressure of pure water at 25oC is 23.8 mmHg.
Solution
Mass of solute = 100g
Mass of solvent = 1000g
Molar mass of solute = 342g/mol
Molar mass of solvent = 18g/mol
Vapor pressure of solvent = 23.8mmHg
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)

Example 4.
The vapor pressure of ether (molar mass 74g/mol) is 442mmHg at 293k. If 3g of compound A are dissolved in 50g of ether at this temperature, the vapor pressure falls to 426mmHg. Calculate the molar mass of A assuming that the solution of A in ether is very dilute.
Solution
Mass of solute = 3g
Mass of solvent = 50g
Vapor pressure of solution P = 426mmHg
V.p of solvent Po = 442mmHg
Mr of solvent = 74g/mol
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)
Example 5.
18. 2g of urea is dissolved in 100g of water at 50oC . The following of vapor pressure produced is 5mmHg. Calculate the molecular mass of urea if the vapor pressure of water at 50oC is 92mmHg.

Solution
Mass of solute = 18.2g
Mass of solvent = 100g
Lowering vapor pressure (Po – P) = 5mmHg
V . p of solvent (Po) = 92mmHg
Required to find Mr of solute
EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)




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EcoleBooks | CHEMISTY As LEVEL(FORM FIVE) NOTES - PHYSICAL CHEMISTRY - Two Components Liquid System(1)

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