Measurement of lowering of vapour pressure

MEASUREMENT OF LOWERING OF VAPOUR PRESSURE

(1) Barometric Method

– Raoult measured the individual vapour pressure of a liquid and then the solution by this method.

– He introduced the liquid or the solution into Toricellian vacuum of a barometer tube and measured the depression of the mercury level.

– This method is neither practicable nor accurate as the lowering of vapour pressure is too small.

(2) Manometric Method

– The vapour pressure of a liquid or solution can be conveniently measured with the help of a manometer (see Fig).

– The bulb B is charged with the liquid or solution. The air in the connecting tube in then removed with a vacuum pump.

– When the stopcock is closed, the pressure inside is due only to the vapour evaporating from the solution or liquid.

– This method is generally used for aqueous solutions.

– The manometric liquid can be mercury or n-butyl phthalate which has low density and low volatility.

(3) Ostwald and Walker’s Dynamic Method (Gas Saturation Method)

– In this method the relative lowering of vapour pressure can be determined straightway.

– The measurement of the individual vapour pressures of a solution and solvent is thus eliminated.

Procedure

The apparatus used by Ostwald and Walker is shown in the following Fig.

– It consists of two sets of bulbs :

(a) Set (A) containing the solution

(b) Set (B) containing the solvent

– Each set is weighed separately.

– A slow stream of dry air is then drawn by suction pump through the two sets of bulbs.

– At the end of the operation, these sets are reweighed. From the loss of weight in each of the two sets, the lowering of vapour pressure is calculated.

– The temperature of the air, the solution and the solvent must be kept constant throughout.

Calculations

– As the air bubbles through set (A) it is saturated up to the vapour pressure p_s of solution and then up to vapour pressure p of solvent in set (B).

– Thus the amount of solvent taken up in set (A) is proportional to p_s and the amount taken up in set (B) is proportional to (p – p_s).

– If w₁ and w₂ be the loss of weight in set A and B respectively,

w₁ ∝ p_s       …(1)

w₂ ∝ p – p_s      …(2)

Adding (1) and (2),we have

w₁ + w₂ ∝ p_s + p – p_s

∝ p      …(3)

Dividing (2) by (3), we can write

(4)

– Knowing the loss of mass in set B (w₂) and the total loss of mass in the two sets (w₁ + w₂), we can find the relative lowering of vapour pressure from equation (4).

– If water is the solvent used, a set of calcium chloride tubes (or a set of bulbs containing conc. H₂SO₄) is attached to the end of the apparatus to catch the escaping water vapour. Thus the gain in mass of the CaCl₂-tubes is equal to (w₁ + w₂), the total loss of mass in sets (A) and (B).

SOLVED PROBLEM

Problem (1): A current of dry air was passed through a solution of 2.64 g of benzoic acid in 30.0 g of ether (C₂H₅OC₂H₅) and then through pure ether. The loss in weight of the solution was 0.645 g and the ether 0.0345 g. What is the molecular mass of benzoic acid ?

solution

According to the theory of Ostwald-Walker method,

From Raoult’s Law, we have

Hence, m, the molecular mass of benzoic acid = 122

Problem (2): A stream of dry air was passed through a bulb containing a solution of 7.50 g of an aromatic compound in 75.0 g of water and through another globe containing pure water. The loss in mass in the first globe was 2.810 g and in the second globe it was 0.054 g. Calculate the molecular mass of the aromatic compound. (Mol mass of water = 18)

solution

According to the theory of Ostwald-Walker method,

Problem (3): In an experiment air was drawn successively through a solution of sugar (38.89 g per 100 g water) and distilled water, and then through anhydrous calcium chloride. It was found that the water lost was 0.0921 g and calcium chloride tubes gained 5.163 g. Find the molecular mass of the sugar. (Mol mass of H2O = 18)

solution

According to the theory of Ostwald-Walker method,

In this case,
w₂, the loss of mass of water = 0.0921 g
(w₁ + w₂), the total loss of solution and solvent = 5.163 g

Substituting values in equation (1)

Hence m, mol mass of sugar = 385