CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – System
CHEMISTRY FORM 5-PHYSICAL CHEMISTRY – System
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).
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
The distribution of solute in a pair of immiscible liquid is governed by the Partition law or Distribution 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.
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
Concentration is the amount of substance per unit volume.
During extraction of solute the amount of solute in extracted component will be decreasing while the amount of extracting component will be increasing.
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.
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.
If layers are not specified then the word “between” shows the numerator and denominator of the formula.
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
Mass of solute X in benzene = 0.13g
Volume of benzene = 10ml
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
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
Volume of water = 50ml
Let X be the weight of succinic acid in water
Conc. of succinic in water = W/50
∴The weight of the acid represent in aqueous solution 1.196g
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?
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
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;
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.
For extracted %
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 )
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”
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.
- b) (ii) Number of extraction = 2
volume of ether ( Vb ) = 50ml.
volume of water ( Vb ) = 1000ml.
kd = 100
Wo = 11g
Aex = Wo – Are
= 11 – 0.3556
∴ The solute extracted is 10.69g.
- a) Explain the principle of solvent extractions.
- b) What is the condition necessary for solvent extraction?
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.
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.
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”.
Mass of solute Wo = 10g
Volume of extracted = 900cc
Volume of extracting Vb = 900cc
Distribution constant Kd = 8
% remained = 1.11/10 X 100 = 11%
ii) % remained by the second girl
∴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.
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 )
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.
i) X extracted
volume of extracting Vb = 100cm3
volume of extracted Va = 100cm3
Distribution constant k = 4
mass of solute X = 8g
∴The fraction of X extraction is
ii) Fraction of X extracted by using two 50cm3
Amount extracted = 8 – 8/9
= 64/9 g
∴ The amount of X extracted by using 2, 50cm3 is 64/9 g.
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
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.
Volume of petrol Va = 1 litre = 1000cm3
Mass of M X = 5g
Distribution constant kd = 0.05
Volume of solvent S Vb = 100cc
∴The mass removed is 4.55g
∴The amount removed is 5 – 0.138g = 4.86g.
SEPARATION OF IMMISCIBLE LIQUID
Immiscible liquids are separated by the process of 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
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
For immiscible liquids , the ratio of their compositions is equal to the ratio of their vapor pressure.
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
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.
Concentration of X in H2O = 6 g/50 cm3
Concentration of X in ethereal = 108/100
K = 9
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.
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.
Po = 5.3kpa
mo = 0.48
Pw = 96 kpa
Mw = 18
mw = 1g
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
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
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.
PB = 1.39 X 104 N/M2
Pw = 8.5 104 N/M2
MB = 157
Mw = 18
mw = 24g
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.
Atmospheric pressure Patm = 1.01 N/M2
PQ = Patm Pw
Pw = 8.77
PQ = 1.01 – 8.77 X 104
Mw = 18
Mass of Q mB = 51g
Mass of W mw = 49
- 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
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.
PB = 119mmHg
PW = 64 mmHg
MB = 1.61g
mw = 1g
Mw = 18g/mol
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.
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
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.
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).
What is colligative properties?
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”
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
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
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.
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.
Mass of solute = 114g
Mass of solvent = 1000g
Lower vapor pressure = 0.11mmHg (Po – P)
Vapor pressure of solvent = 17.54mmHg
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.
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
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.
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
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.
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).
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)
Add the two equations
W1 + W2 = Po – P + P
W1 + W2 = Po
Relative lowering of
But Po – P = W2 and Po = W1 + W2
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
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).
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
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.
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;
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
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
T 2 is the boiling point of solution
T 1 is the boiling point of solution
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.
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
K is called boiling point elevation constant or ebullioscopic constant (k)
Molar elevation constant is the boiling point elevation produced when 1 mole of solute is dissolved in 1kg of the solvent.
a) Define the following
i) Boiling point
ii) Boiling point elevation constant
iii) Ebullioscopic constant
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).
Number of moles of solute nx = 3
Kb = 0.52
Mas of solvent = 1000g
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
Mass of solute mx = 1g
Mass of solvent = 1og
B.P elevation T = 1.2oC
Kb of solvent = 0.52.
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
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
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).
Mass of solute = 0.2g
Mass of solvent = 20g
B .P elevation ΔT = 0.17k
Kb of ether (solvent) = 2.16
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.
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
Freezing point depression is denoted by ΔT
ΔT = T2 – T1
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
The some derivation as in boiling point elevation
Freezing point depression equation;
Kf is freezing point constant or Cryoscopic constant.
Cryoscopic constant is the temperature expressed when the molar weight of solute dissolved in a kg of solvent.
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
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
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 ).
- i) Calculate the molar mass of solute (179.2)
- ii) What is the molecular formula of solute (mf) if its empirical formula is CH2O?
Mass of solute = 0.946g
Mass of solvent = 15g
- 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
Define the following terms
- i) Boiling point.
- ii) Freezing point.
iii) Freezing point depression.
- iv) Cryoscopic constant.
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
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
- i) Concentration or volume of the solvent.
- 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.
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:
- i) Vant- hoff”s first law of osmotic pressure.
- 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
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.
- i) Determine the molecular mass of Y
- ii) Find the molecular formula of Y if kf for C6H6 is 5.12oC kg mol-1
Mass of solute = 7.85g
Mass of solvent = 301g
F . P depression ΔT = 1.05oC
Kf of solvent = 5.12oC kgmol-1
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)
Mass of solute = 35g
Volume of solution = 1dm3
Osmotic pressure π = 5.51
Temperature T = 20oC + 273k
πv = nRT
- a) Define
- i) Osmosis
- ii) Osmotic pressure
- b) Differentiate between osmosis and diffusion
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.
- 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.
- 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 .
- c) Calculate the osmotic pressure of the following solution at 25oC
- i) Sucrose solution of concentration 0.213M (527.72 N/m2)
- ii) A solution of glucose (C6H12O6) made by dissolving 144g/dm3 . (1.1982 x 106 N/m2)
- c) Solution
Temperature T = 25oC + 273
Molarity = 0.213M
(ii) Molar mass of glucose = 180
Concentration = 144g/dm3 = 144000g/m3
= 144000 ÷ 180
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
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).
Mass of solute = 12g
Mass of water = 300g hence volume = 300cm3
Temperature T = 20o + 273
1m = 100cm
1m3 = 1003cm3
X = 300cm3
X = 3 X 10 -4
Calculate the osmotic pressure of 10% glucose solution at 50oC R= 8.314 Jmol-1k -1
Temperature T = 50oC + 273
Mass = 10% glucose
Molar mass of glucose = C6H2O6
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.
Mass of solute = 37.44g
Volume = 1 dm3
Temperature = 37oC + 273k
Osmotic pressure = 1.37 kPa X 1000
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.
Mass of solute A MA =10g
Volume of solvent = 300cm3 = 0.3m3
Osmotic pressure = 375mmHg
1 atm = 760mmHg
X = 375mmHg
X = 375
Temperature T = 25oC + 273k
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.
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
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.
α is the degree of dissociation
g is the Vant hoff’s factor
N is the number of ions
Calculate the N values for the ionization of the following compounds
- i) Al ( OH)3
- ii) FeCl3Ionization equationiii) BaCl2Ionization equation
- iv) NaCl
The degree of dissociation can be expressed terms of
- i) Percentage%
- 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
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)
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
c.c.p = 0.305oC