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 percent of chlorobenzene in the distillate is:
(A) 50
(B) 60
(C) 79
(D) 38.4

Answer 1

Hint: To answer this question, we should first calculate vapour pressure of chlorobenzene. We should then use Raoult’s Law. It states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Complete answer:
-We will calculate weight percent of chlorobenzene by using Rault’s law. According to Raoult's Law the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.
-Vapour pressure of mixture, ${P_m}$ = $7.7 \times {10^4}$ Pa
-Vapour pressure of water, ${P_w}$ = $7 \times {10^4}$ Pa
-Vapour pressure of chloro-benzene, ${P_C}$ = ${P_m} - {P_w}$
                                                                       = $7.7 \times {10^4} - 7 \times {10^4}$
                                                                       = $0.7 \times {10^4}$ Pa
-We should know that Raoult’s law = ${P_a}\alpha {X_a}$
So, from the Raoult’s law, we can now say that:
         ${P_C}\alpha {X_C}$ (1) and ${P_{water}}\alpha {X_{water}}$ (2)
Dividing equation (1) by (2) we get:
        $\dfrac{{{P_C}}}{{{P_{water}}}} = \dfrac{{{X_C}}}{{{X_{water}}}}$
                   = $\dfrac{{\dfrac{{{n_C}}}{{{n_{Total}}}}}}{{\dfrac{{{n_{water}}}}{{{n_{Total}}}}}}$
                   = $\dfrac{{{n_C}}}{{{n_{water}}}}$
Above obtained equation can also be written as:
               $\dfrac{{{P_C}}}{{{P_{water}}}} = \dfrac{{{n_C}}}{{{n_{water}}}} = \dfrac{{wt.chlorobenzene}}{{M.wt.chlorobenzene}} \times \dfrac{{M.wt.water}}{{wt.water}}$
            $\dfrac{{wt.chlorobenzene}}{{wt.water}} = \dfrac{{{P_C} \times M.wt.chlorobenzene}}{{{P_{water}} \times M.wt.water}}$
         = $\dfrac{{0.7 \times {{10}^4} \times 112.5}}{{7 \times {{10}^4} \times 18}}$
         = $\dfrac{{78.75}}{{126}}$
         = 0.625
           Wt. chlorobenzene = 0.625 × wt. water
-We will now calculate the weight percentage of chlorobenzene:
       \[\% \,weight = \dfrac{{weight\,of\,chlorobenzene}}{{Total\,weight}} \times 100\]
                                 = $\dfrac{{0.625}}{{1 + 0.625}} \times 100$
                                 = 38.46%
The mass percentage of chlorobenzene is 38.46%.

Hence the correct option will be: (D) 38.4 .

Note:
We should know that in chemistry; measure the relative proportions of two or more quantities in a mixture. The concentration of a solute is very important in studying chemical reactions because it determines how often molecules collide in solution and thus indirectly determines the rates of reactions and the conditions at equilibrium. We should note that concentration may be expressed in a number of ways. The simplest statement of the concentrations of the components of a mixture is in terms of their percentages by weight or volume. Mixtures of solids or liquids are frequently specified by weight percentage concentrations, such as alloys of metals or mixtures used in cooking, whereas mixtures of gases are usually specified by volume percentages. Very low concentrations may be expressed in parts per million (ppm), as in specifying the relative presence of various substances in the atmosphere.