Question

A mixture of chlorobenzene and water (immiscible) boils at \[90.3{}^\circ C\] at an external pressure of \[740.2{ }mm\]. The vapour pressure of pure water at \[90.3{}^\circ C\] is \[530.1{ }mm\]. Calculate the \[%\] composition of distillate :
A.${{H}_{2}}O=35\%$
B.${{H}_{2}}O=22\%$
C.${{H}_{2}}O=29\%$
D.${{H}_{2}}O=71\%$

Answer 1

Hint:
The percentage composition of any substance could be calculated from the value of partial pressure of individual components, as we can use those values to calculate the mole fraction of the solvent.
 From that we will get the weight of individual components which are involved in the solution, and then we express it in terms of percentage to get the percentage composition.
The mole fraction of a component in the mixture of two components, is the number of moles of that component present, among the total number of moles in the mixture.
Formula used:
${{P}_{solution}}=P_{solvent}^{\circ }{{X}_{solvent}}$
Where,
${{P}_{solution}}$= total vapour pressure of the solution
${{X}_{solvent}}$= mole fraction of the solvent
$P_{solvent}^{\circ }$ = pure vapour pressure of the pure solvent

Complete answer:
Vapour pressure of a substance is the measure of the tendency of that substance to change into its gaseous state and the value of vapour pressure is directly proportional to the temperature.
If we consider the question it is given that at a boiling point vapor pressure of chlorobenzene is ${{p}_{total}}=740.2{ }mm$
And the vapour pressure of water \[{{p}_{{{H}_{2}}O}}=530.1{ }mm\]
Partial pressure of any component in a mixture of components is the pressure exerted by that individual component in the mixture.
As we know total partial pressure is equal to sum of partial pressure of chlorobenzene and partial pressure of water,
${{p}_{chlorobenzene}}=740.2{ }mm-530.1{ }mm=210.1mm$
Mole fraction of any component is the fraction or the number of moles of that component present in the total number of moles of the mixture.
Also, we know that according to Raoult's law, the partial pressure of any substance in a mixture of solution is equal to its mole fraction and the total pressure of the solution. Thus can be mathematically expressed as,
\[{{p}_{chlorobenzene}}={{p}_{total}}\times ~mole{ }fraction\]
Now we will substitute the known values in order to find the mole fraction of the water which is unknown to us,
\[530.1=740.2\times ~mole{ }fraction{ }of~{{H}_{2}}O\]
\[mole{ }fraction{ }of~{{H}_{2}}O=0.716\]
So the mole fraction of water came out to be $0.716$, and the remaining fraction would be the mole fraction of the chlorobenzene since it was a mixture of two components only,
\[mole{ }fraction{ }of~{{C}_{6}}{{H}_{5}}Cl=0.284\]
Now, we Let a \[g~H_2O~\] and b \[g{ }chlorobenzene\] present in distillate,
The mole fraction of water would be,
$\dfrac{\dfrac{{{w}_{{{H}_{2}}O}}}{18}}{\dfrac{{{w}_{chlorobenzene}}}{112.5}+\dfrac{{{w}_{{{H}_{2}}O}}}{18}}=0.716\to (1)$

The mole fraction of chlorobenzene
$\dfrac{\dfrac{{{w}_{chlorobenzene}}}{112.5}}{\dfrac{{{w}_{chlorobenzene}}}{112.5}+\dfrac{{{w}_{{{H}_{2}}O}}}{18}}=0.284\to (2)$
By dividing \[\left( 1 \right)/\left( 2 \right)\]
Where we are assuming, that,
The weight if water ${{w}_{{{H}_{2}}O}}$ is $a$ whereas weight of chlorobenzene ${{w}_{chlorobenzene}}$ is b
$\dfrac{a}{18}\times \dfrac{112.5}{b}=0.403$
$\dfrac{a}{b}=\dfrac{12.89}{31.5}=0.403\to (3)$
Let us assume total distillate mass be $100g$
$a+b=100\to (4)$
Solving both equation $(3)$and $(4)$we get
$b=71.27%$
$a=28.73%$
Hence correct to this question is option “C” which is close to $29\%$.

Additional information:
 Raoult’s law states that a solvent’s partial vapour pressure in a solution (or mixture) is equal or identical to the pure vapour pressure of the pure solvent multiplied by its mole fraction in the solution.
 ${{P}_{solution}}=P_{solvent}^{\circ }{{X}_{solvent}}$
Where,
${{P}_{solution}}$= total vapour pressure of the solution
${{X}_{solvent}}$= mole fraction of the solvent
$P_{solvent}^{\circ }$ = pure vapour pressure of the pure solvent
Consider a solution of pure volatile liquids A and B in a container. Because A and B are both volatile, there would be both particles of A and B in the pure vapour phase.
Hence, the pure vapour particles of both A and B exert partial pressure which tends to the total pressure above the solution.
Raoult’s law can also be applicable to non-ideal solutions. However, this is done by incorporating several factors where we should consider the interactions between molecules of different substances.

Note:The partial pressure of a component in the mixture of two components is the pressure exerted by that individual component in the mixture.
From Raoult’s law it is evident that as the mole fraction of a component reduces, its partial pressure also reduces in the pure vapour phase. Raoult’s law is also similar to the ideal gas law.