Class 12 Chemistry


Vapour Pressure of Solutions of Solids in Liquids

In a pure liquid the entire surface is occupied by molecules of the liquid. In a solution, the surface has molecules of both, i.e. solute and solvent. Hence, the fraction of the surface covered by solvent molecules gets reduced. As a consequence, number of solvent molecules escaping from the surface gets reduced. Thus, vapor pressure is reduced.

The decrease in vapor pressure of solvent depends on the quantity of non-volatile solute present in the solution, irrespective of its nature.

Ideal Solution: A solution which obeys Raoult’s law over the entire range of concentrations is called an ideal solution. The ideal solution has two important properties.

ΔmixH=0 and ΔmixV=0

This means that no heat is absorbed or evolved when the components are mixed to make the solution. Moreover, the volume of solution is equal to the sum of volumes of the two components.

At molecular level, the ideal behavior of the solution can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will be of types A-A and B-B. In the solution, A-B type of interactions will also be present apart from these two interactions. If the intermolecular attractive forces between A-A and B-B are nearly equal to those between A-B, an ideal solution is formed.

Non-ideal Solution: When a solution does not obey Raoult’s law over the entire range of concentrations, it is called non-ideal solution. The vapor pressure of a non-ideal solution can be higher or lower than predicted by Raoult’s law equation. Positive deviation means higher vapor pressure, while negative deviation means lower vapor pressure. When A-B interactions are weaker than A-A or B-B interactions, there is positive deviation. When A-B interactions are stronger than A-A or B-B interactions, there is negative deviation.

Azeotropes: A binary mixure which has the same composition in liquid and vapor phase, and which boils at a constant temperature is called azeotrope. Components of azeotrope cannot be separated by fractional distillation. There are two types of azeotropes, which are as follows:

  1. Minimum boiling azeotrope: The solution which shows a large positive deviation from Raoult’s law forms minimum boiling azeotrope at a specific composition. Example: ethanol-water mixture
  2. Maximum boiling azeotrope: The solution which shows a large negative deviation from Raoult’s law forms maximum boiling azeotrope at a specific composition. Example: nitric acid-water mixture.

Colligative Properties: Some properties of a solution depend on the number of solute particles relative to the total number of particles present in a solution. Such properties are called colligative properties. Various colligative properties are as follows:

Relative Lowering of Vapor Pressure:

The relation between vapor pressure of solution, mole fraction and vapor pressure of solvent is given by following equation:


Now, reduction of vapor pressure of solvent (Δp1) can be given as follows:


`=p_1^0(1-x_1)` …………………(1)

We know that x2 `= 1-x_1`

So, Δ`p_1=x_2p_1^0` ………….. (2)

Above equation can be written as follows:


Or, `(p_1^0-p_1)/(p_1^0)=x_2`

It is clear from above equation that relative lowering of vapor pressure is equal to the mole fraction of the solute. Above equation can also be written as follows:


For dilute solutions n2 < < n1

Hence, neglecting n2 in denominator, we get


Or, `(p_1^0-p_1)/(p_1^0)=(w_2xx\M_1)/(M_2xx\w_1)`

Here, w1 and 22 are the masses and M1 and M2 are the molar masses of the solvent and solute respectively.

Example: The vpaor pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol-1). Vapor pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?

Answer: Given: `p_1^0=0.850` bar, `p=0.845` bar, `M_1=78` g mol-1, `w_2=0.5` g and `w_1=39` g

Molar mass of solid substance can be calcualted as follows:


`=(0.850 text(bar)-0.845text(bar))/(0.850 text(bar))``=(0.5g\xx78g\text(mol)^(-1))/(M_2xx39g)`

Or, `M_2xx0.005=(0.850xx0.5xx78)/(39)`

Or, `M_2=(0.850xx0.5xx78)/(39xx0.005)=170` g mol-1