Question

The vapour pressure of a certain pure liquid A at 100 K is 40 mbar. When a solution of B is prepared in A at the same temperature, the vapour pressure is found to be 32 mbar. The mole fraction of A in the solution is:
a.) 0.5
b.) 0.2
c.) 0.1
d.) 0.8

Answer 1

Hint: In this question, the total pressure of the solution has decreased when A is added to B, there B is a non-volatile solute. We will assume that this solution is dilute and use the colligative properties.

Complete step by step solution:
This question deals with colligative properties. Let us first look into colligative properties and the factors on which they depend.
Colligative properties of solutions arise due to the presence of non-volatile solute in a solution. They depend on the number of the solute particles and not on the nature of the chemical species present in a solution. When a non-volatile solute is added to a solvent, it displaces some solvent molecules at the surface of the solution thereby reducing vapour pressure of the solvent which is called the relative lowering of vapour pressure.
We can apply the calculations for colligative properties only for the dilute solutions since their behaviour is very close to that of an ideal solution.
Relative lowering of vapour pressure:
Vapour pressure is exhibited by the vapours present above the surface of a liquid at a particular temperature such that the vapour phase and the liquid phase are in equilibrium with each other. When a non-volatile solute is added to a solvent, the total vapour pressure of the solution decreases.
According to Roult’s law, the total vapour pressure of a solution at a given temperature is given by:
$ { P }_{ T }=\sum { { P }_{ i }^{ \ast }{ x }_{ i } } $
Where $ { P }_{ i }^{ \ast }$ is the vapour pressure of the component that is present in the solution without the presence of other components of the solution (i.e. it is present in its pure state without the interference of any other component) and $ { x }_{ i }$ is the mole fraction of that component in the solution.
For a solution with a solvent (A) and a non-volatile solute (B), the same equation will be written as:

${ P }_{ T }={ P }_{ A }^{ \ast }{ x }_{ A }$ ….(1) (the component B will not contribute towards the total vapour pressure since it is non-volatile)
Since, $ { P }_{ T }$=32 mbar, $ { P }_{ A }^{ \ast }$=40 mbar, we need to find ${ x }_{ A }$. We will substitute these values in equation (1):
${ x }_{ A }=\cfrac { 32\quad mbar }{ 40\quad mbar } =0.8$

So, the correct answer is “Option D”.

Note: If the non-volatile solute dissociates in the solvent, then the number of particles of the solute will increase. Since Colligative properties depend only on the number of the particles, the van ’t Hoff factor is used in order to account for the new number of particles of the solute.