Question

Mole fraction of component A in vapour phase is${\mathrm{x}}_1$​ and mole fraction of component A in liquid mixture is ${\mathrm{x}}_2$​, then, the total vapour pressure of liquid mixture is :

Answer 1

Step 1: Understand the term “Mole fraction”

• The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all the components of the solution.

$\mathrm{Mole}\mathrm{fraction}\mathrm{of}\mathrm{solute}=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{\mathrm{Moles}\mathrm{of}\mathrm{solute}+\mathrm{Moles}\mathrm{of}\mathrm{solvent}}$

• $\mathrm{Mole}\mathrm{fraction}\mathrm{of}\mathrm{b}=\frac{{\mathrm{n}}_{\mathrm{b}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}}$
• ${\mathrm{n}}_{\mathrm{b}}$= number of moles of solute
• ${\mathrm{n}}_{\mathrm{a}}$ = Number of moles of solvent

Step 2: Analyze the question statment.

• Mole fraction of component A in vapour phase is ${\mathrm{x}}_1$​. So, using Dalton's law of partial pressure:
  For component A ,

${\mathrm{P}}_{\mathrm{A}}={\mathrm{P}}_{\mathrm{T}}\times {\mathrm{x}}_1$​

• Mole fraction of compound A in liquid mixture is ${\mathrm{x}}_2$,
• Now applying Raoult’s law which states that for solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
• ${\mathrm{P}}_{\mathrm{A}}={\mathrm{x}}_2\times {{\mathrm{P}}_{\mathrm{A}}}^0$

Step 3 : Comparing the two equations

• So, from both equations we get

${\mathrm{x}}_2\times {{\mathrm{P}}_{\mathrm{A}}}^0={\mathrm{x}}_1\times {\mathrm{P}}_{\mathrm{T}}$

• So, ${\mathrm{P}}_{\mathrm{total}}=\frac{{\mathrm{x}}_2}{{\mathrm{x}}_1}\times {{\mathrm{P}}_{\mathrm{A}}}^0$