Question

The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of the water is 0.9, the vapour pressure of the solution will be:
A. 30 mm of Hg
B. 24 mm of Hg
C. 21 mm of Hg
D. 27 mm of Hg

Answer 1

Hint:In order to answer this question, you must enlist all the concepts of Solutions and go through all the Raoult’s laws and must use the concept of vapour pressure. For solution, firstly evaluate the mole fraction of the component from the given data. And then calculate the vapour pressure of the solution, and then the vapour pressure of the solution is the required quantity. And this will give you the required answer.

Complete step by step answer:
Step 1: In this step we will enlist all the given quantities:
The given quantities are:
Vapour Pressure of Water at room temperature = 30 mm of Hg
Mole fraction of water = 0.9
Vapour pressure of the solution
Step 2: In this step we will calculate the vapour pressure of the solution using mole fraction of Water:
${P_A}^ \circ $ = 30 mm of Hg
Mole fraction of water = x = $0.9$
Vapour pressure of solution, ${P_A}\, = \,{P_A}^ \circ {X_A}$
Vapour pressure = ${P_A}$ = $30 \times 0.9$ = $27$ mm of Hg
Hence we got the required value of Vapour Pressure.
So, the correct answer is option D.

Note: Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature. Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. For a solution of two liquids A and B, Raoult's law predicts that if no other gases are present, then the, total pressure above the solution of A and B would be
${P_S}\, = \,{X_A}{P_A}^ \circ \, + \,{X_B}{P_B}^ \circ $