Question

A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be?
A.0.549
B. 0.200
C. 0.786
D.0.478

Answer 1

Hint: In order to answer this question, you must enlist all the concepts of Solutions and go through all the laws and must use the concept of total pressure. For solution, firstly evaluate the mole fraction of each component from the given ratio. And then calculate the total pressure of the solution and then calculate the fraction of pentane in the vapour phase. 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:
Ratio of mole fraction of pentane and hexane = 1: 4
Vapour Pressure of Pentane = 440 m of Hg
Vapour Pressure of Hexane = 120 mm of Hg
Temperature = 20°C

Step 2: In this step we will calculate the individual mole fraction of pentane and hexane:
Mole fraction of Pentane solution = $\frac{1}{{1 + 4}}$ = $\frac{1}{5}$
Mole fraction of Hexane solution = $\frac{4}{{1 + 4}}$ = $\frac{4}{5}$

Step 3: In this step, we will calculate the total pressure of the solution:
Total pressure of solution, ${P_S}\, = \,{X_p}{P_p}^ \circ \, + \,{X_h}{P_h}^ \circ $
$ \Rightarrow \,{P_S}\, = \,\frac{1}{5} \times 440\, + \,\frac{4}{5} \times 120\,\, = \,184\,mm\,Hg$

Step 4: In this step we will find the mole fraction of pentane in vapour phase:
Mole fraction of pentane in vapour phase $ = \,\frac{{{X_p}{P_p}^ \circ }}{{{P_S}}}\, = \,\frac{{88}}{{184}}\, = \,0.478$
Hence we got the required value of mole fraction.

Hence the correct answer is option D.

Note: 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 $