Question

Liquids A and B form an ideal solution in the entire composition range. At \[350{\text{ K}}\], the vapour pressures of pure A and pure B are \[7 \times {10^3}{\text{ Pa}}\]and \[12 \times {10^3}{\text{ Pa}}\], respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is:
A.\[{{\text{x}}_{\text{A}}} = 0.37,{{\text{x}}_{\text{B}}} = 0.63\]
B.\[{{\text{x}}_{\text{A}}} = 0.28,{{\text{x}}_{\text{B}}} = 0.72\]
C.\[{{\text{x}}_{\text{A}}} = 0.76,{{\text{x}}_{\text{B}}} = 0.24\]
D.\[{{\text{x}}_{\text{A}}} = 0.4,{{\text{x}}_{\text{B}}} = 0.6\]

Answer 1

Hint: The pure vapor pressure of gas A and gas B are given to us. Using Dalton's law of partial pressure and the relationship between partial pressure and mole fraction, we can calculate the number of moles in vapor phase.
Formula used: \[{{\text{Y}}_{\text{A}}} = \dfrac{{{{\text{P}}_{\text{A}}}}}{{{{\text{P}}_{\text{T}}}}}\] here \[{{\text{Y}}_{\text{A}}}\] is mole fraction of A in vapor phase, \[{{\text{P}}_{\text{A}}}\] is partial pressure of A, \[{{\text{P}}_{\text{T}}}\] is total pressure of mixture.
\[{{\text{P}}_{\text{A}}} = {\text{P}}_{\text{A}}^{\text{o}} \times {\chi _{\text{A}}}\] here \[{\text{P}}_{\text{A}}^{\text{o}}\] is pure vapor pressure of liquid A and \[{\chi _{\text{A}}}\] is mole fraction of A. \[{{\text{P}}_{\text{T}}} = {{\text{P}}_{\text{A}}} + {{\text{P}}_{\text{B}}}\]
\[{\text{mole fraction}} = \dfrac{{{\text{mole percentage}}}}{{100}}\]

Complete step by step answer:
Mole percent of A is given that is 40 percent and hence,
\[{\chi _{\text{A}}} = \dfrac{{40}}{{100}} = 0.4\]
A fraction is always calculated from 1 or the sum of the fraction of all the components in a mixture is always 1. Hence we have mole fraction of B as \[1 - 0.4 = 0.6\]
Now we will calculate the partial pressure of A using the formula, we have been given pure vapor pressure of A that is \[7 \times {10^3}{\text{ Pa}}\].
\[{{\text{P}}_{\text{A}}} = 7 \times {10^3}{\text{ Pa}} \times 0.4 = 2.8 \times {10^3}\]
We will calculate the partial pressure of B using the formula, we have been given pure vapor pressure of A that is \[12 \times {10^3}{\text{ Pa}}\] and mole fraction of B is \[0.6\].
\[{{\text{P}}_{\text{B}}} = 12 \times {10^3}{\text{ Pa}} \times 0.6 = 7.2 \times {10^3}\]
Using the formula we will now finally calculate the number of moles of A in vapor state:
\[{{\text{Y}}_{\text{A}}} = \dfrac{{2.8 \times {{10}^3}}}{{2.8 \times {{10}^3} \times 7.2 \times {{10}^3}}}\]
\[ \Rightarrow {{\text{Y}}_{\text{A}}} = \dfrac{{2.8 \times {{10}^3}}}{{10 \times {{10}^3}}} = 0.28\]
Hence the mole fraction of A is \[0.28\]. The mole fraction of B will be \[1 - 0.28 = 0.72\]

So the correct option is B.

Note:
Dalton’s law of partial pressure says that the total pressure of a mixture of gases is equal to the sum of pressure of each gas when considered separately. An ideal solution is a solution that follows all laws under all conditions of temperature and pressure.