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
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
2. 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
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 BOILING POINT 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
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
3. 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
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
= 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 ).
i) Calculate the molar mass of solute (179.2)
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
F. 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
i) Boiling point.
ii) Freezing point.
iii) Freezing point depression.
iv) Cryoscopic constant.
Answers
i) Boiling point Is the temperature st which liquid boils where by the vapor pressure of that liquid is equal to the atmospheric pressure.
ii) Freezing point Is the temperature at which liquid change into solid state. e.g f.P of water is OoC.
iii) Freezing point depression Is the difference between the freezing point of the solution and that of the solvent (pure liquid).
iv) Cryoscopic constant Is 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.
ii) Freezing point Is the temperature at which liquid change into solid state. e.g f.P of water is OoC.
iii) Freezing point depression Is the difference between the freezing point of the solution and that of the solvent (pure liquid).
iv) Cryoscopic constant Is 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
4. 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
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.
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:
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
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.
i) Determine the molecular mass of Y
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
a) Define
i) Osmosis
ii) Osmotic pressure
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.
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.
ii. 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
= 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
Solution
Mass of solute = 12g
Mass of water = 300g hence volume = 300cm3
Temperature T = 20o + 273
= 293k
= 293k
But
1m = 100cm
1m = 100cm
1m3 = 1003cm3
X = 300cm3
1003
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
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.
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
i) Al ( OH)3
Ionization equation
ii) FeCl3
Ionization equation
iii) BaCl2
Ionization equation
iv) NaCl
Ionization equation
NOTE
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
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
ms X Mrx
= 1.86 x 2.4 x 1000
250 x 58.5
250 x 58.5
= 0.305oC
c.c.p = 0.305oC
7 Comments