# Colligative Properties

The properties of dilute solutions which depend only on number of solute particles of solute present in the solution and not on their identity are called colligative properties (denoting depending upon collection).

We shall assume here that the solute is non volatile, so it does not contribute to the vapour. We shall also assume that the solute does not dissolve in solid solvent, that is, the pure solid solvent separates when the solution is frozen, the latter assumption is quite drastic, although it is true for many mixtures, it can be avoided.

Colligative properties are the properties of only dilute solutions which are supposed to behave as ideal solutions. The various colligative properties are:

(A) Lowering of vapour pressure

(B) Osmotic pressure

(C) Elevation of boiling point

(D) Depression of freezing point

#### Lowering of Vapour Pressure by a Non-Volatile Solute

When a non-volatile solute is added to a solvent it lowers the vapour pressure of a solvent. It is because the solute molecules block the part of surface and hence reduce the rate at which solvent molecules escape from the solution.
However the solute molecules have no effect on rate at which solvent molecules return because returning molecules can stick to the any part of surface even if solute particle is there. (*Remember solute is soluble in solvent through some attractive forces) .

At equilibrium the rate of evaporation equals the rate of condensation. The vapour pressure, therefore, is lower than in pure solvent.

In 1886, the French chemist, Francois Raoult, after a series of experiments on a number of solvents including water, benzene and ether, succeeded in establishing a relationship between the lowering of vapour pressure of a solution and the mole fraction of the non-volatile solute.

Let us consider a solution obtained by dissolving n moles of a non-volatile solute in N moles of a volatile solvent.
Then mole fraction of the solvent, X1 = N/(n+N)

and mole fraction of the solute, X2 = n /(N +n).

Since the solute is non-volatile, it would have negligible vapour pressure. The vapour pressure of the solution is, therefore merely the vapour pressure of the solvent. According to Raoult’s law, the vapour pressure of a solvent (p1) in an ideal solution is given by the expression

P1 = X11 ……….(i)

where P°1 is the vapour pressure of the pure solvent.

Since X1 + X2 = 1 , Eq. 1 may be written as

P1 = (1 – X2) P°1 ……………(ii)

P1/P°1 = (1 – X2)

X2 = (P°1 – P1)/P°1 …………(iii)

The expression on the left hand side of Equation (iii) is usually called the relative lowering of vapour pressure.
Equation (iii) may thus be stated as:
‘ The relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute present in the solution .’

This is one of the statements of the Raoult’s law.

Since mole fraction of the solute, X2 is given by n/(N+n), Equation (iii) may be expressed as

$\large \frac{P_1^o – P_1}{P_1^o} = \frac{n}{N+n}$    ……(iv)

It is evident from Equation (iv) that the lowering of vapour pressure of a solution depends upon the number of moles (and hence on the number of molecules) of the solute and not upon the nature of the solute dissolved in a given amount of the solvent. Hence, lowering of vapour pressure is a colligative property.

##### Determination of Molar Masses from Lowering of Vapour Pressure

It is possible to calculate molar masses of non-volatile non-electrolytic solutes by measuring vapour pressures of their dilute solutions.
Suppose, a given mass, w gram, of a solute of molar mass m, dissolved in W gram of solvent of molar mass M, lowers the vapour pressure from P°1 to P1

Then , by Equation (iv)

$\large \frac{P_1^o – P_1}{P_1^o} = \frac{w/m}{W/M + w/m}$ ……(v)

Since in dilute solutions, n is very small as compared to N, Equation (v) may be put in the approximate form as

$\large \frac{P_1^o – P_1}{P_1^o} = \frac{n}{N} = \frac{w/m}{W/M} = \frac{w M}{W m}$ ….(vi)

$\large m = \frac{w M}{W(P_1^o – P_1)/P_1^o}$

##### Limitations of Raoult’s Law :

(i) Raoult’s law is applicable only to very dilute solutions.

(ii) It is applicable to solutions containing non-volatile solute only.

(iii) It is not applicable to solutes which dissociate or associate in a particular solution.

Illustration : Calculate the mass of the solute of molar mass 342 g mol-1 that should be dissolved in 150 gram of water to reduce its vapour pressure to 22.8 torr. The vapour pressure of pure water at 25°C = 23.75 torr.

Solution: we know,

$\large \frac{P_1^o – P_1}{P_1^o}= \frac{w_2 M_1}{w_1 M_2}$

Substituting the various values in the above equation ,

$\large \frac{23.75 torr – 22.8 torr}{23.75 torr}= \frac{w_2 \times 18 g \; mol^{-1}}{150 g \times 342 g\; mol^{-1}}$

∴ w2 = 114 g

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