Question

At $40^{o} C$, the vapour pressure of pure liquids, benzene and toluene, are 160 mm Hg and 60 mm Hg, respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution, should be:
(A) 140 mm Hg
(B) 110 mm Hg
(C) 220 mm Hg
(D) 100 mm Hg

Answer 1

Hint: Use the Raoult’s law, which states that for a solution of volatile liquids the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. The proportionality constant would vapour pressure of the pure component.

Complete step by step solution:
The Raoult’s law, which was given by the French chemist, Francois Marte Raoult (1886), gives us the quantitative relationship between vapour pressure and mole fraction of the respective solution of volatile liquids
The Raoult’s law states that for a solution of volatile liquids the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
$\implies p_i \propto x_i $
$\implies p_i = x_i \times p_{i}^{o} $
Where $p_i$ is the vapor pressure of the component in the given solution
$p_{i}^{o}$ is the vapor pressure of pure component
$\implies p_{total} = p_{1}^{o}x_1 + p_{2}^{o}x_2 $
Given equimolar solution
$\implies x_1 = x_2 = 0.5$
$\implies p_{total} $ = 160 x 0.5 + 60 x 0.5
=110 mm Hg

So, the correct answer is option (B) which is 110 mm Hg

Additional information:
The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero.

Note: In real life, a solution need not obey Raoult’s law over the entire range of concentration. If the vapour pressure is higher, the solution exhibits positive deviation, and if the vapour pressure is lower, the solution exhibits a positive deviation. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions, whereas the solutions which do obey Raoult’s law over the entire range of concentration are known as non-ideal solutions.