Question

The mole fraction of \[A\]vapours in the above solution in mixture of $A$ and $B$(${\chi _A} = 0.4$) will be:
$[P_A^\circ = 100mm;P_B^\circ = 200mm]$
a)$0.4$
b)$0.8$
c)$0.25$
d) None of the above

Answer 1

Hint:The vapour pressure of liquid in liquid solution can be calculated by using Raoult’s law of volatile solute which states that for a solution of volatile liquids the partial pressure of each component is directly proportional to its mole fraction.
Formula used: ${P_A} = {\chi _A} \times P_A^\circ $
${P_T} = {P_A} + {P_B}$And ${P_A} = {y_A} \times {P_T}$

Complete step by step answer:
Vapour pressure is defined as the pressure exerted by the vapours above the liquid surface in equilibrium with the liquid phase at a given temperature.
Raoult’s law for volatile solute states that for a solution of volatile liquids the partial pressure of each component is directly proportional to its mole fraction. The relation is represented by:
${P_A}\alpha {\chi _A}$
$ \Rightarrow {P_A} = {\chi _A} \times P_A^\circ $ Where ${P_A}$ the partial vapour pressure of is $A$,${\chi _A}$ is the mole fraction of $A$in solution and $P_A^\circ $ is the vapour pressure of pure component$A$.
Now the relation for total partial vapour pressure can be derived as-
 \[{P_T} = {P_A} + {P_B}\] Where ${P_A}$ and ${P_B}$ are the partial vapour pressure of component $A$ and $B$.\[ \Rightarrow {P_T} = {P_A}^\circ \times {\chi _A} + {P_B}^\circ \times {\chi _B}\]
As we know the total mole fraction of a solution is always equal to one.
  $ \Rightarrow{\chi _A} + {\chi _B} = 1$
 $ \Rightarrow {\chi _B} = 1 - {\chi _A} $
It is given in the question ${\chi _A} = 0.4$ in the solution, so we can calculate the value of ${\chi _B}$.
  $ \Rightarrow {\chi _B} = 1 - {\chi _A} $
  $ \Rightarrow {\chi _B} = 1 - 0.4 $
  $ \Rightarrow {\chi _B} = 0.6 $
Substituting the values of mole fraction to find the total pressure, it is mentioned in the question that $P_A^\circ = 100mm;P_B^\circ = 200mm$
$ \Rightarrow {P_T} = 100mm \times 0.4 + 200mm \times 0.6 $
$ \Rightarrow {P_T} = 40mm + 120mm $
$ \Rightarrow {P_T} = 160mm$
To calculate the mole fraction of liquid in vapour phase we have to substitute the above values in following the equation: ${P_A} = {y_A} \times {P_T}$
$ \Rightarrow 40 = {y_A} \times 160 $
$ \Rightarrow {y_A} = \dfrac{{40}}{{160}} $
$ \Rightarrow {y_A} = 0.25 $
Hence the mole fraction of component $A$ in vapour phase is equal to $0.25$.

So,Correct option is (c).
Note:
 The composition of the vapour phase in equilibrium with the solution can be determined from the partial pressure of the two components i.e. ${P_A} = {y_A} \times {P_T}$
The mole fraction in the solution of a component like ${\chi _A}$should not be misinterpreted with the mole fraction of component in the vapour phase like ${y_A}$ as it needs the total partial pressure for its calculation.