SOLUTION APPENDIX-II

CLASS –XII

TOPIC COVERED – VAPOUR PRESSURE, RAOULT’S LAW AND AZEOTRPES

Vapour Pressure of pure liquid and solution

Vapour Pressure: When a liquid is taken in a closed vessel, a part of the liquid evaporates and its vapour pressure
occupy the available empty space. These vapours cannot escape as the vessel is closed. They would rather have a
tendency to condense into liquid form. In fact an equilibrium is established between vapour phase and liquid phase
and the pressure that its vapour exert is termed as vapour pressure. Thus, vapour pressure of the liquid may be
defined as the pressure exerted by the vapours above the liquid surface in equilibrium with the liquid phase at
a given temperature.

The vapour pressure of a liquids depends on the following factors:

Nature of the liquid: Liquid having weak intermolecular attraction are volatile and therefore have great vapour
pressure.

Temperature: Vapour pressure of a liquid increases with increase in temperature. This is because with increase in
temperature, the kinetic energy of the molecules increases and therefore large number of molecules are available for
escaping from the surface of the liquid.

(a) Vapour pressure of liquid – liquid solution:

Raoult’s Law for solution of volatile liquid: It states that for a solution of volatile liquids the partial pressure of
each component solution is directly proportional to its mole fraction, Mathematically,

PA ∝ xA PB ∝ xB

PA = 𝑃𝐴0 xA PB = 𝑃𝐵0 xB

Where PA and PB are partial pressure, xA and xB are the mole fractions, 𝑃𝐴0 and 𝑃𝐵0 are the vapour pressure of pure
component A and B respectively.

If PS is the total pressure then according to Dalton’s law of partial pressure,

P = PA + PB

= 𝑃𝐴0 xA + 𝑃𝐵0 xB

= 𝑃𝐴0 (1-xA) + 𝑃𝐵0 xB

= 𝑃𝐴0 + (𝑃𝐴0 - 𝑃𝐵0 ) xB

As 𝑃𝐴0 and 𝑃𝐵0 are constants at a given temperature it is evident from the above equations that the total vapour pressure
varies linearly with the mole fraction xB (or xA since xA = 1 – xB).

The composition of the vapour phase in equilibrium with the solution can be determined from the partial pressure of
the two components. If yA and yB are the mole fraction of A and B respectively in the vapour phase, then

PA = yA Ptotal(PS) and PA = yB Ptotal(PS)

(b) Vapour pressure of solutions of solids in liquids:

Raoult’s Law for a solution containing a non-volatile solute and volatile solvent: It states that the relative
lowering of vapour pressure is equal to mole fraction of solute which is non – volatile.

Mathematically, PS = PA + PB
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 PS = PA(Since solute B is non - volatile)

PS = 𝑃0𝐴 xA

PS = 𝑃0𝐴 (1 – xB) = 𝑃0𝐴 - 𝑃0𝐴 xB

𝑃0𝐴 xB = 𝑃0𝐴 - PS
0
𝑝𝐴 −𝑃
Or = 𝑥𝐵
𝑃𝐴0
Or Relative lowering of vapour pressure = Mole fraction of solute
Ideal and non- ideal solutions:

(a) Ideal solution: A solution is called an ideal solution it obeys Raoult’s

PS = PA + PB = 𝑃0𝐴 xA + 𝑃0𝐵 xB

Liquids having similar nature and structure are likely to from ideal
solutions. Examples are:

 Mixture of methanol and ethanol

Reason for the formation of ideal solutions: A solution of two miscible liquids A and B will be ideal if two essential
conditions are fulfilled.

(i) If FA – A is the force of attraction between molecules of A and FB – B is that of molecules of B, then A and B will
form an ideal solution only if,

FA – B = FA – A = FB – B

(ii) The solution of A and B liquids will be ideal if A and B have similar structures and polarity. Methanol and
ethanol have the same functional group and almost same polarity and therefore, form ideal solutions.

For an ideal solution:

(i) Raolut’s Law is obyed.

(ii) ∆mixH = 0 and (iii) ∆mixV = 0

(b) Non-ideal solution: A solution which does not obey Raolut’s Law for all concentrations is called a non – Ideal
solution.

For a non – Ideal solution

(i) Raoult’s Law is not obeyed, i.e., PA ≠ 𝑃0𝐴 xA and PB ≠ 𝑃0𝐵 xB

(ii) (ii) ∆mixH ≠ 0 and (iii) ∆mixV ≠ 0solvent

A non – ideal solution can show either positive or negative deviation from Raoult’s law.

(A) Positive Deviation: The deviation will be called positive when the partial pressure of each component and the
resultant total pressure are greater than the pressure expected on the basis of Raoult’s Law.

In such cases, the intermolecular forces between solvent – solute molecules (FA – B) are weaker than those between
solvent – solvent (FA – A) and solute – solute (FB – B) molecules. That is,

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 FA – B < FA – A and FB – B

This shows that the molecules of A or B will escape more easily from the surface of the solution, i.e., the vapour
pressure of solution will be higher.

Characteristics of solution showing positive deviation

 PA > 𝑃𝐴0 xA ; PB > 𝑃𝐵0 xB

Some examples of the solution exhibiting positive deviation are:

(i) Ethyl alcohol and water

(ii) Acetone and carbon tetra chloride

(iii) Carbon tetrachloride and benzene

(iv) Acetone and benzene

(B) Negative Deviation: The deviation is called negative deviation, if the partial pressure of each component (A
and B) and resultant total vapour pressure are less than the pressure expected on the basis of Raoult’s Law.

This type of deviation is shown by the solution in which:

FA – B > FA – A and FB – B

Due to this there is decrease in the escaping tendency of A and B molecules from the surface of solution.
Consequently, the vapour pressure of the solution will be lower.
Characteristic of solution showing negative deviation:

 PA < 𝑃𝐴0 xA ; PB < 𝑃𝐵0 xB

Some examples of the solution showing negative deviation are:

(i) HNO3 and water

(ii) Chloroform and acetone

(iii) Hydrochloric acid and water

Azeotropes or Azeotropic mixture: Azeotropes are binary mixture having the same composition in liquid and
vapour phase and boil at a temperature.

Types of Azeotropes:

(i) Minimum boiling azeotropes: These are the binary mixtures whose boiling s less than either of the two
components. The non-ideal solutions which show a large positive deviaton from Raoult’s law from minimum
boiling azeotrope at a specific composition, eg., a mixture of 94.5% ethyl alcohol and 4.5% water by volume.

(ii) Maximum boiling azeotropes: These are the binary mixtures whose boiling point is more than either of the
two components. The solutions that show large negative from Raoult’s law from maximum boiling azeotrope at
specific composition, eg., a mixture of 68% HNO3 and 32% H2O by mass.

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 QUESTION BASED ON NOTES:

1. Vapour pressure of water at 293 K is 17.51 mm. Lowering of vapour pressure of a sugar solution is 0.0614 mm.
calculate

i. Relative lowering of vapour pressure [0.00351]

2. The vapour pressure of a 5% aqueous solution of a non – volatile organic substance at 373 K is 745 mm. calculate
the molar mass of the solute. [48 g 𝑚𝑜𝑙 −1 and 47.05 g 𝑚𝑜𝑙 −1]

3. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non – volatile, non – electrolyte solid
weighing 0.5 g is added to 39.0 g of benzene (molar mass 78 g 𝑚𝑜𝑙 −1 ). The vapour pressure of the solution is
0.845 bar. What is the molar mass of the solid substance? [170 g 𝑚𝑜𝑙 −1 ]

4. At 298 K, the vapour pressure of water is 23.75 mm Hg. Calculate the vapour pressure at the same temperature
over 5% aqueous solution of urea [CO(NH2)2]. [23.375 mm]

5. A current of dry air was passed through a solution of 2.5 g of a non volatile substance ‘X’ in 100 g of water and
then through water alone. The loss of weight of the former was 1.25 and that of the latter was 0.05 g. calculate

i. mole fraction of the solution [0.0385]

6. A solution containing 6 g of benzoic acid in 50 g of ether (C2H5OC2H5) has a vapour pressure of 410 mm of
mercury at 293 K. given that the vapour pressure of ether at the same temperature is 442 mm of mercury, calculate
the molecular mass of benzoic acid. (Assume that the solution is dilute). [122.65 u]

7. Calculate the vapour pressure at 295 K of a 0.1 M solution of urea. The density of the solution may be taken as 1
g/cm3. The vapour pressure of pure water at 295 K is 20 mm. [19.96 mm]

8. Vapour pressure of an solution of glucose is 750 mm of Hg at 373 K. calculate the molality and mole fraction of
solution. [𝑥2 = 0.013, molality = 0.73 m]

9. How much urea (molar mass = 60 g 𝑚𝑜𝑙 −1) should be dissolved in 50 g of water so that its vapour pressure at
room temperature is reduced by 25%. Calculate molality of the solution obtained. [55.556 g, 18.52 m]

10. At 500C, the vapour pressure of pure CS2 is 854 torr. A solution of 2.0 g of sulphur in 100 g of CS2 has vapour
pressure of 848.9 torr. Determine the formula of sulphur molecule. [S8]

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