Question

a. The vapour pressure of n-hexane and n-heptane at$\text{273 K}$ are $\text{45}\text{.5 mmHg}$ and $\text{11}\text{.4 mmHg}$, respectively. What is the composition of a solution of these two liquids if the vapour pressure of the liquids at $\text{273 K}$ is $\text{37}\text{.3 mmHg}$.
b.The mole fraction of n-hexane in the vapour of a solution of n-hexane and n-heptane is $0.75$ at $\text{273 K}$. What is the composition of the liquid solution?
A.$\left( \text{a} \right)\text{ 0}\text{.76, 0}\text{.24 }\left( \text{b} \right)\text{0}\text{.61}$
B.$\left( \text{a} \right)\text{ 0}\text{.67, 0}\text{.33 }\left( \text{b} \right)\text{0}\text{.57}$
C.$\left( \text{a} \right)\text{ 0}\text{.59, 0}\text{.41 }\left( \text{b} \right)\text{0}\text{.67}$
D.None of these

Answer 1

This problem is based on Raoult's law which states that the partial pressure of each component in an ideal solution is the partial pressure of the pure solvent multiplied by the mole fraction of the solute in the solution.

Formula used: The Raoult’s law can be expressed mathematically as,
${{\text{p}}_{\text{i}}}\text{= }{{\text{p}}_{\text{0}}}\text{ }{{\text{x}}_{\text{i}}}$
where ${{\text{p}}_{\text{i}}}$the partial pressure of the solution, the vapour pressure of the pure solvent is${{\text{p}}_{\text{0}}}$and${{\text{x}}_{\text{i}}}$is the mole fraction of the solute in the solution.

Complete step by step answer:
Given that the vapour pressure of n-hexane and n-heptane at $\text{273 K}$ are $\text{45}\text{.5 mmHg}$ and $\text{11}\text{.4 mmHg}$, respectively.
Total vapour pressure after they were mixed at$\text{273 K}$is$\text{37}\text{.3 mmHg}$
According to the Raoult’s Law: ${{\text{p}}_{1}}\text{= p}_{1}^{0}\text{ }{{\text{x}}_{1}}$ where $\text{p}_{1}^{0}$ is the vapour pressure of n-hexane,${{\text{x}}_{1}}$is the mole fraction of the same.
Again,${{\text{p}}_{2}}\text{= p}_{2}^{0}\text{ }{{\text{x}}_{2}}$, where $\text{p}_{2}^{0}$ is the vapour pressure of n-heptane,${{\text{x}}_{2}}$ is the mole fraction of the same.
$\text{p}_{\text{1}}^{\text{0}}\text{= 45}\text{.5 mmHg}$, $\text{p}_{\text{2}}^{\text{0}}\text{ =11}\text{.4 mmHg}$
We know that, the total mole fractions of the solute and the solvent is equal to 1, hence,
${{\text{x}}_{\text{1}}}\text{+ }{{\text{x}}_{\text{2}}}=1\Rightarrow {{\text{x}}_{\text{2}}}\text{=1 - }{{\text{x}}_{\text{1}}}$.
Total pressure
${{\text{p}}_{\text{1}}}\text{+ }{{\text{p}}_{\text{2}}}$
= \text{ p}_{1}^{0}\text{ }{{\text{x}}_{1}}+\text{p}_{2}^{0}\text{ }{{\text{x}}_{2}}=\text{p}_{1}^{0}\text{ }{{\text{x}}_{1}}+\text{p}_{2}^{0}\text{ }\left( 1-{{\text{x}}_{1}} \right)$$$$=\text{p}_{1}^{0}\text{ }{{\text{x}}_{1}}+\text{p}_{2}^{0}\text{ - p}_{2}^{0}{{\text{x}}_{1}}={{\text{x}}_{1}}\left( \text{p}_{1}^{0}-\text{p}_{2}^{0} \right)+\text{p}_{2}^{0}
Putting the values in the above equation we get,
${{\text{x}}_{1}}\left( \text{45}\text{.5}-\text{11}\text{.4 } \right)+\text{11}\text{.4 }=\text{37}\text{.3 }$
$\Rightarrow {{\text{x}}_{1}}=\dfrac{\text{37}\text{.3 -11}\text{.4}}{34.1}=0.76$
Hence, ${{\text{x}}_{\text{2}}}\text{=1 - 0}\text{.76 = 0}\text{.24}$
The mole fractions of n-hexane and n-heptane in the solutions are $\text{ 0}\text{.76 and 0}\text{.24 }$ respectively.
(b) When the mole fraction of n-hexane in the vapour of a solution of n-hexane and n-heptane is $0.75$at$\text{273 K}$, we have that, $\dfrac{{{\text{x}}_{\text{1}}}\left( \text{45}\text{.5} \right)}{\text{37}\text{.3}}\text{=0}\text{.75}$.
Hence, we know that, ${{\text{x}}_{\text{1}}}\text{=}\dfrac{\text{0}\text{.75}\times \text{37}\text{.3}}{\left( \text{45}\text{.5} \right)}$=$0.61$

Hence, the correct answer is option C.

Note:
There are certain limitations to the application of the Raoult’s law, as this law is applicable only to the ideal solutions in which the solute molecules neither dissociate nor do they associate in the solution. So the van’t Hoff factor for these solutions is equal to unity.