Form 5 Chemistry – PHYSICAL CHEMISTRY – Three Components Liquid System

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 P^o is be the vapor pressure of solvent

             P be the vapor pressure of solution

W₁ α  P

_{W1 α  P}⁰ _{– P}

W₁ = KP  But k =1

W₁ = p ————(i)

W₂ = k (P^o -P)    k =1

W₂ = P^o – P ———(ii)

 Now

Add the two equations

W₁ + W₂ = P^o – P + P

W₁ + W₂ = P^o

From Raoults,

Relative lowering of

But   P^o – P = W₂  and  P^o = W₁ + W₂

^Hence;

 

Where

P^o   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

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
m_s mass of solvent in kg

n_x number of moles of solute

If m_s 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 n_x = 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 1.2^oC. Calculate the molar mass of solute if the ebullioscopic constant of water is 0.52

Solution

Mass of solute m_x = 1g

Mass of solvent = 1og

B.P elevation T = 1.2^oC

Kb  of solvent = 0.52.

      

Example 3.

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

Kb for 0.52^oC kgmol^-1

Solution

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

Mass of solute m_s = 100g

B.P of solution T₂ = 101.04^oC

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

Solution

B .P of solvent T₁ = 56.38^oC

B.P of solution T₂ = 56.88^oC

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

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

Solution

F.P of ethanoic T₁ = 16. 63^oC

Mass of solute = 2. 5g

Mass of solvent = 40g

Lowered  F.P T₂ = 11.48^oC

K_f of the solvent = 3.9^oC kgmol^-1

                   ΔT = T₂ – T₁
= 11.48 – 16.63^o

                       = -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 -0.651^oC ( k_f  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 = -0.651^oC

K_f of solvent = 1.86

F.P of solvent = 0^oC

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 O^oC.
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 1.05^oC below that of pure benzene.

    i) Determine the molecular mass Y.

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