Question

A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressure of pure hydrocarbons at ${{20}^{o}}C$ is 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in vapour phase would be:
(A) 0.786
(B) 0.478
(C) 0.549
(D) 0.200

Answer 1

Generally, liquid solvents are volatile. The solute may be or may not be volatile. For solutions of volatile liquids, the partial pressure of each component of this solution is directly proportional to its mole fraction present in the solution. This relationship is known as Raoult’s Law.

Complete step by step solution:
Consider two components 1 and 2 in the solution. ${{x}_{1}},{{x}_{2}}\And {{p}_{1}},{{p}_{2}}$ are the mole fraction and partial pressure of the two components respectively.
According to Raoult’s law, for component 1,
${{p}_{1}}\alpha {{x}_{1}}$ and, ${{p}_{1}}={{p}^{o}}_{1}{{x}_{1}}$
For component 2, ${{p}_{2}}\alpha {{x}_{2}}$ and ${{p}_{2}}={{p}^{o}}_{2}{{x}_{2}}$
Where ${{p}_{1}}^{0}\And {{p}_{2}}^{o}$ are the vapour pressure of the pure component 1 and 2 respectively.
Then the total pressure, $\begin{aligned} & {{p}_{total}}={{p}_{1}}+{{p}_{2}} \\\ & {{p}_{total}}={{x}_{1}}{{p}_{1}}^{o}+{{x}_{2}}{{p}_{2}}^{o} \\\ \end{aligned}$ --- equation (1)
Given component 1 =pentane and component 2 = hexane
The mole ratio of pentane to hexane = 1:4
Then, the mole fraction of pentane, ${{x}_{1}}=\dfrac{1}{1+4}=\dfrac{1}{5}$
The mole fraction of hexane, ${{x}_{2}}=\dfrac{4}{1+4}=\dfrac{4}{5}$
Given, the vapour pressure of pure pentane, ${{p}_{1}}^{o}$ = 440 mm of Hg
Vapour pressure of pure hexane, ${{p}_{2}}^{o}$ = 120 mm of Hg
Substitute the above values in equation (1),
Then, the total pressure of the solution, $\begin{aligned} & {{p}_{total}}={{x}_{1}}{{p}_{1}}^{o}+{{x}_{2}}{{p}_{2}}^{o} \\\ & {{p}_{total}}=\dfrac{1}{5}X440+\dfrac{4}{5}X120=184mmofHg \\\ \end{aligned}$
The mole fraction of pentane in vapour phase = $\dfrac{{{x}_{1}}{{p}^{o}}_{1}}{{{p}_{total}}}=\dfrac{(\dfrac{1}{5}X440)}{184}=0.478$

Hence, the correct answer is option B.

Note: The conclusion from the equation of total pressure of the solution is, total vapour pressure over the solution can be related to the mole fraction of any compound and varies linearly with the mole fraction of component 2.