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CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – System

CHEMISTRY FORM 5-PHYSICAL CHEMISTRY - System

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CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – 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




The immiscible pair is kept in a separating funnel in order to see them clearly.

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

 

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.

In terms of Extractions

Note

Concentration is the amount of substance per unit volume.
Also

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

Kd = 15

Let the amount extracted be ‘a’

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

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;

∴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

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

For more than one extraction we normally use the following formula;

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 = WoAre

 

If the amount extracted and amount remained are known, then their respective percentages can be calculated.
i.e
For extracted %

 

Example 1

Solute the partition law

Calculate the amount of solute extracted by shaking 1 litre of aqueous solution containing 11g of Q with;

  1. i) 100ml of ether ( 10g)
  2. 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

10x = 1100 – 100x

110x = 1100

x = 10g.

∴The solute extracted is 10g.

  1. b) (ii) Number of extraction = 2

volume of ether ( Vb ) = 50ml.

volume of water ( Vb ) = 1000ml.

kd = 100

Wo = 11g

= 0.30556

Again

Aex = Wo – Are

= 11 – 0.3556

= 10.69g

∴  The solute extracted is 10.69g.




Example 2

  1. a) Explain the principle of solvent extractions.
  2. 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 = 900cc

Volume of extracting Vb = 900cc

Distribution constant Kd = 8

= 1.11g

% remained = 1.11/10 X 100 = 11%

ii) % remained by the second girl

= 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

∴The fraction of X extraction is

ii) Fraction of X extracted by using two 50cm3

= 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.”

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)

∴The mass removed is 4.55g

= 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

For immiscible liquids , the ratio of their compositions is equal to the ratio of their vapor pressure.

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

K = 9
Now

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

Note

Unit conversion

i) 1 N/m2 = 1 pa

ii) 1 atm = 760 mmHg

iii) 1 atm = 1.01 105 Pa

iv) 1 atm = 1.01 105 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 = 8.5 104 N/M2

MB = 157

Mw = 18

mw = 24g

from




Example 4.

An organic liquid Q which do not mix with water distills in steam at 96oC under the pressure of 1.01 . The pressure of water at 96oC is 8.77 N/M-2 . The distillate contains 51% by mass Q . Calculate the molar mass of Q.

Solution

Atmospheric pressure Patm = 1.01 N/M2

PQ = Patm Pw

Pw = 8.77

Now

PQ = 1.01 – 8.77 X 104

= 1033

MQ= ?

Mw = 18

Mass of Q mB = 51g

Mass of W mw = 49

From
Example 5.

  1. i) Define the term steam distillation.
  2. 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
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

= 760 733

Pn = 27

Mw = 18

Mn = ( 126) + 5+ 14 ( 16 2)

Mn = 123

From
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

 

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

 

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

For mole fraction of solute

Let n be number of mole solute

N be number of mole solvent

The molar mass of solute can be determined

 

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

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

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

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

 

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

 

EFFECT OF SOLUTE ON VAPOR PRESSURE BY OSWALD’S METHOD

Consider the solution made by dissolving solute in a given solvent. Then the dry air being passed through the two component

The passage of dry air cause the loss in mass (weight).

Let:

W1  be loss in mass of solution.

W2  be loss in mass of solvent.

Also the Po is be the vapor pressure of solvent

P be the vapor pressure of solution

W1 α  P

W1 α  P0 – P

W1 = KP  But k =1

W1 = p ————(i)

W2 = k (Po -P)    k =1

W2 = Po – P ———(ii)

Now

Add the two equations

W1 + W2 = Po – P + P

W1 + W2 = Po

From Raoults,

Relative lowering of

But   Po – P = W2  and  Po = W1 + W2

Hence;

 

Where

Po   is the V.p of solvent

P  is the V.p of solution

W1  is the loss in mass of solution

W2  is the loss in mass of solvent




Example 6

A current of dry air was passed through a solution of 2.64g of benzoic acid in 30g of ether (C2H5OC2H5) and then through pure ether. The loss in weight of the solution was 0.64g and that of ether was 0.0345g. Calculate the molecular mass of benzoic acid (122g/mol).

Solution

Weight of solution W1 = 0. 645g

Weight of solvent W2 = 0.0345g

Mass of solution m = 2.64g

Mass of solvent M = 30g

Mr of ether (C2H5OC2H5) = 74g/mol

But
Example 7

A stream of dry air was passed through a bulb containing a solution of 7.5g of aromatic compound in 75cm3 of water and through another globe containing pure water. The loss in mass in the first globe was 2.81g and in the second globe was 0.054g. Calculate the Molar mass of aromatic compound 93.6.

Solution

Mass of solution = 7.5g

Mass of solvent = 75g

Loss in mass of solution = W1 = 2.81g

Loss in mass of solvent = W2 = 0.054g

From Oswald’s law;

Example 8

In a experiment air was drown successively through a solution of sugar (38.89g per 100g H2O) and the distilled water, and then through anhydrous calcium chloride. It was found that water lost was 0.0921g and the calcium chloride globe gained 110g. Calculate the molar mass of sugar

Solution

Mass of sugar = 38.89g, W1 + W2 = 5.16g

Mass of water 100g,  W2= 0.0921g

 

BOILING POINT ELEVATION

What is boiling point?

Boiling point is the temperature at which liquid boils where by the vapor pressure of that liquid is equal to the atmospheric pressure.

Effect of solute(impurities) to the boiling point of the liquid.

When solute particles are added to the liquid, the solution formed. The boiling point of the solution formed is increased by some oC. The boiling point is elevated due to the increase in collision of solute molecules and liquid molecules. Finally the temperature being raised.

The difference between the boiling point of the solution and the boiling point of the liquid is what we call Boiling point elevation

The boiling point is denoted by ΔT

ΔT = T2 – T1

Where by;

T 2 is the boiling point of solution

T 1 is the boiling point of solution

Note

ΔT is always positive.

The relationship between boiling point elevation and the amount of solute added

This relationship can be explained by two laws which are;
i) Blagden’s law
ii) Raoult’s law

 

i) BLAGDEN’S LAW

It states that;

“The change in temperature caused by addition of solute is directly proportional to the amount of solute being added”

If the amount is represented by m

From Blagden’s law

 

ii) RAOULT’S LAW OF BOILINGPOINT ELEVATION

It state that;

“The change in temperature caused by the addition of solute particles is inversely proportional the molecular weight of the solute added”

From Raoult’s law

 

What is Molality?

Molality is the number of moles of solute per 1kg of the solvent.

 

Where
ms mass of solvent in kg

nx number of moles of solute

If ms is given in ‘g’

Then it has to be converted to kg

But

K is called boiling point elevation constant or ebullioscopic constant (k)

Definition

Molar elevation constant is the boiling point elevation produced when 1 mole of solute is dissolved in 1kg of the solvent.
Example 1

a) Define the following

i) Boiling point

ii) Boiling point elevation constant

iii) Ebullioscopic constant

Answer

i) Boiling point is the temperature at which liquid boils where by the vapor pressure of that liquid is equal to the atmosphere pressure.

ii) Boiling point elevation constant is the temperature change when 1 mole of solute is dissolved 1kg of the solvent.

iii) Ebullioscopic constant is the boiling point elevation obtained when 1 mole of solute is dissolved in 1kg of the solvent.

b) What is the boiling point of the solution containing 3moles of sugar in 1000g of water (Kb for H2O is 0.52).

Solution

Number of moles of solute nx = 3

Kb = 0.52

Mas of solvent = 1000g

Example 2

When 1g of solute was added in 10g of water the boiling point elevation was 1.2oC. Calculate the molar mass of solute if the ebullioscopic constant of water is 0.52

Solution

Mass of solute mx = 1g

Mass of solvent = 1og

B.P elevation T = 1.2oC

Kb  of solvent = 0.52.




Example 3.

A solution containing 18.4g glycerine per 100g of water boil at 101.04oC. Calculate the molar mass of glycerine .

Kb for 0.52oC kgmol-1

Solution

Mass of solute mx = 18.4g (T1 = B.P of H2O)

Mass of solute ms = 100g

B.P of solution T2 = 101.04oC

Kb of solvent = 0.52

Example 4

The boiling point of a solution containing 0.2g of substance X in 20g of ether is 0.17k higher than that of the pure ether . Calculate the molecular mass of  X . The boiling point constant of ether per 1kg is 2.16k (127. 8g/mol).

Solution

Mass of solute = 0.2g

Mass of solvent = 20g

B .P elevation  ΔT = 0.17k

Kb of ether (solvent) = 2.16

Example 5

Acetone boils at 56.38oC and a solution of 1.41g of an organic solid in 20g of acetone boils at 56.88oC . If k for acetone per 100g is 16.7oC . Calculate the mass of one mole of the organic solid.

Solution

B .P of solvent T1 = 56.38oC

B.P of solution T2 = 56.88oC

Mass of solute = 1.41g

Mass of solvent = 20g

Kb of solvent = 1.67

 

FREEZING POINT DEPRESSION

What is freezing point?

Is the temperature at which liquid change into solid state /freeze.

Effect of solute on the freezing property of the liquid

When solute is added to a certain liquid, the freezing point of the liquid is lowered. This is because the solute particles disturb the intermolecular forces between the molecules of the liquids.

The difference between the freezing point of the solvent (pure liquid) and that of solution is what we call freezing point

Depression

Freezing point depression is denoted by ΔT

ΔT = T2 – T1

Where;

T1 is the freezing point of solvent

T2 is the freezing point of solution

The freezing point depression (T) is related to the amount of solute added by the following expression

Note

The some derivation as in boiling point elevation

Definition

Freezing point depression equation;

Kf is freezing point constant or Cryoscopic constant.

Definition

Cryoscopic constant is the temperature expressed when the molar weight of solute dissolved in a kg of solvent.

Example 1

Ethanoic acid has the freezing point of 16.63oC on adding 2.5g of solute to 40g of the acid, The freezing point was lowered to 11.48oC. Calculate the molar mass of the solute if kf is 3.9oC kgmol-1

Solution

F.P of ethanoic T1 = 16. 63oC

Mass of solute = 2. 5g

Mass of solvent = 40g

Lowered  F.P T2 = 11.48oC

Kf of the solvent = 3.9oC kgmol-1

ΔT = T2 – T1
= 11.48 – 16.63o

= -5.150C

Now

 

Example 2

When 0.946g of organic substance was added in 15g of water resulting into the solution which was found to have the freezing point of -0.651oC ( kf  is 3.9 ).

  1. i) Calculate the molar mass of solute (179.2)
  2. ii) What is the molecular formula of solute (mf) if its empirical formula is CH2O?

Solution

Mass of solute = 0.946g

Mass of solvent = 15g

  1. P of solution = -0.651oC

Kf of solvent = 1.86

F.P of solvent = 0oC

T = T2 T1

T = 0.651 – 0

T = -0.651

Example 3

Define the following terms

  1. i) Boiling point.
  2. ii) Freezing point.

iii) Freezing point depression.

  1. iv) Cryoscopic constant.

Answers

i) Boiling pointIs the temperature st which liquid boils where by the vapor pressure of that liquid is equal to the atmospheric pressure.
ii) Freezing pointIs the temperature at which liquid change into solid state. e.g f.P of water is OoC.
iii) Freezing point depressionIs the difference between the freezing point of the solution and that of the solvent (pure liquid).
iv) Cryoscopic constantIs the temperature depressed when the molar weight of solute is dissolved in a kilogram of solvent.

b) 7.85g of the compound Y having the empirical formula of C5H4 was dissolved in 310g of benzene (C6H6) . If the freezing point of the solution is 1.05oC below that of pure benzene.

i) Determine the molecular mass Y.

ii) Find the molecular formula of Y if  Kf  for C6H6 is 5.12oC kg mol-1




OSMOTIC PRESSURE

What is osmotic pressure ?

Osmotic pressure: Is the force per unit area which occur as the results of solvent to flow From low concentration to the high concentration of solute through a semi permeable membrane .

Osmotic pressure causes the osmosis to take place.

Osmosis: Is the tendency of solvent molecular to migrate from the region of low concentration to the region of high concentration of solute.

Consider the simple experiment below

 

Factors affecting the Osmotic pressure of the substance

The Osmotic pressure of the substance affected by the two factors

  1.   i) Concentration or volume of the solvent.
  2. ii) Temperature

 

i) CONCENTRATION OR VOLUME OF THE SOLVENT

The osmotic pressure of the substance varies directly proportional to the concentration and inversely proportional to the volume .

i.e An increase in concentration of the solution will cause an increase in osmotic pressure of the solution.

Also increases in osmotic pressure cause the decrease in volume.

 

ii) TEMPERATURE

The temperature varies directly proportional to the osmotic pressure as long as the temperature do not exceed the optimum temperature of the semi permeable membrane.

The effect of two factors above are explained by Vant- hoff’s laws of osmotic pressure.

 

VANT – HOFF’S LAWS OF OSMOTIC PRESSURE

Vant – hoff tried to put forward his laws which explain the variation of osmotic pressure with concentration and temperature.

There are two laws developed, These are:

  1. i) Vant- hoff”s first law of osmotic pressure.
  2. ii) Vant – hoff’s second laws of osmotic pressure.

 

i) VANT – HOFF’S FIRST LAW OF OSMOTIC PRESSURE

It states that;

“The osmotic pressure of the solution is inversely proportional to its volume”

Let the osmotic pressure be denoted by π

From 1st law

 

ii) VANT – HOFF’S SECOND LAW OF OSMOTIC PRESSURE

It states that;

“The osmotic pressure of the solution is directly proportional to temperature”

From 2nd law

This is for 1 mole, But for ‘n’ mole the formula will be

 

Example1

7.85g of the compound Y having the empirical formula of C5H4 was dissolved in 301g of Benzoic (C6H6) . If the freezing point of the solution is 1.05oC below that of pure benzene.

  1. i) Determine the molecular mass of  Y
  2. ii) Find the molecular formula of  Y if kf for C6H6 is 5.12oC kg mol-1

Solution

Mass of solute = 7.85g

Mass of solvent = 301g

F . P depression ΔT = 1.05oC

Kf of solvent = 5.12oC kgmol-1

Example 2

Calculate the molar mass of solute given that 35g of solute dissolved in 1dm3 has osmotic pressure of 5.51 N/M2 at 20oC. (R = 8.314 Jmol-1k-1)

Solution

Mass of solute = 35g

Volume of solution = 1dm3

Osmotic pressure π = 5.51

Temperature T = 20oC + 273k

= 293k

πv = nRT

Example 3

  1. a) Define
  2. i) Osmosis
  3. ii) Osmotic pressure
  4. b) Differentiate between osmosis and diffusion

Answers
a) i. Osmosis : Is the tendency of a solvent molecules to move from the region of law concentration to the region of high concentration of solute.

  1. Osmotic pressure : Is the force per unit area which occur as the result of solvent to flow from low concentration to the high concentration of solute through a membrane.
  2. b) Osmosis is the movement of solvent molecules from the region of low concentration to the region of high concentration of solute through a semi- permeable membrane while Diffusion is the movement of gases molecules from the region of high concentration to a region of low concentration .
  3. c) Calculate the osmotic pressure of the following solution at 25oC
  4. i) Sucrose solution of concentration 0.213M (527.72 N/m2)
  5. ii) A solution of glucose (C6H12O6) made by dissolving 144g/dm3 . (1.1982 x 106 N/m2)
  6. c) Solution

Temperature T = 25oC + 273

= 298k

Molarity = 0.213M

(ii) Molar mass of glucose = 180

Concentration = 144g/dm3 = 144000g/m3

= 144000 ÷ 180

= 800M

Osmotic pressure π = MRT

= 800 x 8.314 X 298

= 1.982 X 106N/M2

∴The osmotic pressure of a solution = 1.982 X 106 N/M2

 

Example 4

Calculate the osmotic pressure of the solution containing 12g of C6H12O6 in 300g of water at 20oC.

R = 8. 314 Jmol -1k-1(54.13 X 102 N/m-2).

Solution

Mass of solute = 12g

Mass of water = 300g  hence volume = 300cm3

Temperature T = 20o + 273
= 293k

But
1m = 100cm

1m3 = 1003cm3

X = 300cm3
1003

X = 3 X 10 -4

Example 5

Calculate the osmotic pressure of 10% glucose solution at 50oC  R= 8.314 Jmol-1k -1

Solution

Temperature T = 50oC + 273

= 323k

Mass = 10% glucose

Molar mass of glucose = C6H2O6

= 180




Example 6

37.44g of haemoglobin, the protein of red blood cell which carries the oxygen in the blood, were dissolved in a dm3 of water.

The solution formed had an osmotic pressure of 1.37 kPa  at body temperature of 37oC.

Determine the molar mass of haemoglobin.

Solution

Mass of solute = 37.44g

Volume = 1 dm3

Temperature = 37oC + 273k

=310k

Osmotic pressure = 1.37 kPa X 1000

= 1370Pa

Example 7

A solution containing 10g of solute A in 300. Of water has an osmotic pressure of 375mm of 25oC. What will be the osmotic pressure be dissolving 2.55g of solute A in 50g of water? R = 0.0821 atm mol-1 k -1 L.

Solution

Mass of solute A MA =10g

Volume of solvent = 300cm3 = 0.3m3

Osmotic pressure = 375mmHg

1 atm = 760mmHg

X = 375mmHg

X =  375
760

= 0.49

Temperature T = 25oC + 273k

= 298k

 

COLLIGATIVE PROPERTIES OF ASSOCIATIVE AND DISSOCIATIVE SOLUTE

Colligative properties depends on vant hoff’s factor .Vant  hoff  factor is the ratio of observed colligative properties to that of calculated colligative property.

OR

Vant hoff factor is the ratio of experimental colligative property to that of theoretical colligative property.

 

VANT HOFF FACTOR

Vant hoff’s factor is denoted by ‘g’ or ‘l’ and therefore

Vant’s hoff  factor is obtained when the solute added is not completely ionized (dissociated).
Also some solutes, when added in the solvent tend to associate.

The phenomena of associating or dissociating of solute is explained in terms of degree (degree of association or dissociation).

The degree of dissociation is

Define:

The degree of dissociation is the percentage or fraction of moles of ions which have gone into the solution.

It is denoted by α

The degree of dissociation is related to Vant hoff’s factor and the number of ions formed by the solute dissolved.

Where by;

α is the degree of dissociation

g is the Vant hoff’s factor

N is the number of ions

Example 1.

Calculate the N values  for the ionization of the following compounds

Solution

  1. i) Al ( OH)3

Ionization equation

  1. ii) FeCl3Ionization equationiii) BaCl2Ionization equation
  2. iv) NaCl

Ionization equation

 

NOTE

The degree of dissociation can be expressed terms of

  1. i) Percentage%
  2. ii) Decimals

When it is expressed in percentage it cannot exceed 100% and

When it is expressed in decimal it cannot exceed 1.00

Also when solute is ionizing completely its degree of dissociation become 100%  OR  1.00

Example 2.

What will be the boiling point of the solution containing 2.4g NaCl in 250cm3 of water .it an aqueous solution of NaCl is 70% Dissociation.

(kb water = 1.86)

Solution

Mass of solute mx = 2.4g.

Mass of solvent ms = 250g.

Kb = 1. 86.

α = Degree of elevation 70% or o.7http://192.168.137.101/tz/cexam.php?MASTexam=586

Mrx = NaCl

= 23 + 35.5

= 58.5 g/mol

Δ T = Kb x m x 1000
ms X Mrx

= 1.86 x 2.4 x 1000
250 x 58.5

= 0.305oC

c.c.p = 0.305oC

RELATED POST;

CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – Solutions

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CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – Gases



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