Question

Water and chlorobenzene are immiscible liquids. Their mixture boils at $89^\circ C$ under a reduced pressure of $7.7 \times {10^4}Pa$ . The vapour pressure of pure water at $89^\circ C$ is $7 \times {10^4}Pa$ . Weight percentage of chlorobenzene in the distillate is:
A. $50$
B. $60$
C. $79$
D. $38.46$

Answer 1

We have to find the partial pressure of the Chlorobenzene and apply the raoult’s law to find out the mole fraction, using mole fraction find out the weight percentage.
According to Raoult’s law, if the partial pressure is decreased, the mole fraction also gets decreased.
Raoult’s Law, ${{\text{P}}_{\text{A}}} \propto {\chi _{\text{A}}}$
Where,
${\chi _A}$ is mole fraction of a component A
${P_A}$ is partial pressure of component A

Complete step by step answer:
Let’s start the question by writing the things that are given to us in this question. We are provided with the following date,
The temperature at which the mixture is being boiled (T) = 89oC
The pressure of at which boiling is achieved (P) = $7.7 \times {10^4}Pa$
The vapour pressure of Pure water at 89oC (Pw) = $7 \times {10^4}Pa$
We need to find the weight percentage of the chlorobenzene in the distillate and for that we need to find the mole fraction of chlorobenzene. Mole fraction is the number of mole of compound A present in total number of moles of solution i.e. A+B. For finding the mole fraction we will use the Raoult's law, and for that first we need to find the pressure of chlorobenzene.
The pressure of chlorobenzene (PChloro) = $P-{P_w} = 7.7{\text{ }} \times {10^4}Pa-7 \times {10^4}Pa = 0.7 \times {10^4}Pa$
So, according to raoult's law,${{\text{P}}_{\text{A}}} \propto {\chi _{\text{A}}}$, This means
$\dfrac{{{{\text{P}}_{{\text{Chloro}}}}}}{{{{\text{P}}_{{\text{Water}}}}}} = {\text{ }}\dfrac{{{\chi _{{\text{Chloro}}}}}}{{{\chi _{{\text{Water}}}}}}$, This gives us the ratio of mole fractions which will be equal to
$\dfrac{{{\chi _{{\text{Chloro}}}}}}{{{\chi _{{\text{Water }}}}}} = \dfrac{{0.7}}{7} = \dfrac{1}{{10}}$, This means that 1 Chlorobenzene and 10 water is there in mixture.
The molecular mass of Chlorobenzene is $112.56g/mol$, if $11$ moles are there only $112.56g/mol$of chlorobenzene will be distillate and the water will not evaporate. The total mass of 11 mole of mixture is $292.57g$ and $112.56g$ is evaporated which means the weight percentage is $38.46$ .

So, the correct answer is Option D.

Note: We must know that the Raoult’s Law is one of the most famous laws which is used in chemistry and related fields. Raoult's law states that the partial pressure of one component in the liquid is always proportional to the mole fraction of the compound. Raoult’s law has been used for calculating the mole fraction from partial pressures or the partial pressure ratio from the mole fractions. These data helps in various experiments.