Question

Vapour pressure of chloroform $\left( {{\text{CHC}}{{\text{l}}_{\text{3}}}} \right)$ and dichloromethane $\left( {{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_2}} \right)$ at ${\text{298 K}}$ are ${\text{2}}00{\text{ mm Hg}}$ and ${\text{415 mm Hg}}$ respectively.
(i) Calculate the vapour pressure of the solution prepared by mixing ${\text{25}}{\text{.5 g}}$ of chloroform and ${\text{40 g}}$ of dichloromethane at ${\text{298 K}}$ and
(ii) mole fraction of each component in solution.

Answer 1

In a binary mixture of two components, the total vapour pressure of the mixture is equal to the sum of the partial pressures of individual components. The partial vapour pressure of each component is the product of the vapour pressure of the pure component and its mole fraction in the mixture.
$P = \left( {P_A^o \times {X_A}} \right) + \left( {P_B^o \times {X_B}} \right) \\\\$
Complete step by step answer:
(i) The vapour pressure of pure chloroform is ${\text{P}}_A^o{\text{ = 2}}00{\text{ mm Hg}}$.
The vapour pressure of pure chloroform is ${\text{P}}_B^o{\text{ = 415 mm Hg}}$.
The molecular weight of chloroform is ${{\text{M}}_A}{\text{ = 119}}{\text{.5 g}}$.
Mass of chloroform is ${\text{25}}{\text{.5 g}}$.
Divide mass of chloroform with molar mass of chloroform to obtain number of moles of chloroform.
$\dfrac{{{\text{25}}{\text{.5 g}}}}{{{\text{119}}{\text{.5 g/mol}}}} = 0.213{\text{ mol}}$
The molecular weight of dichloromethane is ${{\text{M}}_B}{\text{ = 85 g}}$.
Mass of dichloromethane is ${\text{40 g}}$.
Divide mass of dichloromethane with molar mass of dichloromethane to obtain a number of moles of chloroform.
$\dfrac{{{\text{40 g}}}}{{{\text{85 g/mol}}}} = 0.47{\text{ mol}}$
Calculate the mole fraction of chloroform by dividing the number of moles of chloroform with the sum of the number of moles of chloroform and dichloromethane.

= 0.322 \\\ $$ The sum of the mole fractions of chloroform and dichloromethane is equal to one. Calculate the mole fraction of dichloromethane by subtracting the mole fraction of chloroform from one. $${X_B} = 1 - {X_A} \\\ = 1 - 0.322 \\\ = 0.688 \\\\$$ Calculate the total pressure of the solution: $$P = \left( {P_A^o \times {X_A}} \right) + \left( {P_B^o \times {X_B}} \right) \\\ = \left( {200 \times 0.322} \right) + \left( {415 \times 0.688} \right) \\\ = 64.4 + 285.52 \\\ = 349.92 \\\\$$ Hence, the vapour pressure of the solution is $$349.92{\text{ mm Hg}}$$. (ii) The mole fraction of chloroform in the solution is 0.322. The mole fraction of dichloromethane in the solution is 0.688. **Note:** In a binary solution, the mole fraction of a particular component in the vapour phase can be obtained by dividing the partial pressure of that component with the total vapour pressure.