
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
575.1k+ views
Hint: 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, $\begin{align}
& {{p}_{total}}={{p}_{1}}+{{p}_{2}} \\
& {{p}_{total}}={{x}_{1}}{{p}_{1}}^{o}+{{x}_{2}}{{p}_{2}}^{o} \\
\end{align}$ --- 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, $\begin{align}
& {{p}_{total}}={{x}_{1}}{{p}_{1}}^{o}+{{x}_{2}}{{p}_{2}}^{o} \\
& {{p}_{total}}=\dfrac{1}{5}X440+\dfrac{4}{5}X120=184mmofHg \\
\end{align}$
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.
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, $\begin{align}
& {{p}_{total}}={{p}_{1}}+{{p}_{2}} \\
& {{p}_{total}}={{x}_{1}}{{p}_{1}}^{o}+{{x}_{2}}{{p}_{2}}^{o} \\
\end{align}$ --- 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, $\begin{align}
& {{p}_{total}}={{x}_{1}}{{p}_{1}}^{o}+{{x}_{2}}{{p}_{2}}^{o} \\
& {{p}_{total}}=\dfrac{1}{5}X440+\dfrac{4}{5}X120=184mmofHg \\
\end{align}$
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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

State the laws of reflection of light

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

