Question

The vapor pressure of a liquid solution containing A and B is 99 torr. Calculate mole % of B in vapor phase.

(Given :${\displaystyle P_A^0}$ = 100 torr , ${\displaystyle P_B^0}$ = 80 torr)

Answer 1

Step 1: Given information:

• The vapor pressure of the liquid is 90 torr.
• ${\displaystyle P_A^0}$=100 torr
• ${\displaystyle P_B^0}$=80 torr

Step 2: Formula used:

Raoult's law:

• The partial vapor pressure of a component in a mixture is the multiplication of vapor pressure to the mole fractions of pure components.
• $$\begin{gathered} \mathrm{P}=\mathrm{P}{°}_{\mathrm{A}}{\mathrm{x}}_A \\ \mathrm{P}=\mathrm{P}{°}_{\mathrm{B}}{\mathrm{x}}_{\mathrm{B}} \end{gathered}$$
• where P_A and P_B are the partial vapor pressures of components A and B and xA and xB are the mole fractions of components A and B.
• On rearranging, the formula can be written as:

$\mathrm{P}=\mathrm{P}{°}_{\mathrm{A}}+{\mathrm{x}}_{\mathrm{B}}(\mathrm{P}{°}_{\mathrm{B}}-\mathrm{P}{°}_{\mathrm{A}})$-----------------(1)

Step 3: Calculation for mole fraction of B:

On substituting the values in equation (1)

$$\begin{gathered} \mathrm{P}=\mathrm{P}{°}_{\mathrm{A}}+{\mathrm{x}}_{\mathrm{B}}(\mathrm{P}{°}_{\mathrm{B}}-\mathrm{P}{°}_{\mathrm{A}}) \\ 99\mathrm{torr}=100\mathrm{torr}+{\mathrm{x}}_{\mathrm{B}}(80-100)\mathrm{torr} \\ 99\mathrm{torr}=100\mathrm{torr}-20{\mathrm{x}}_{\mathrm{B}}\mathrm{torr} \\ 20{\mathrm{x}}_{\mathrm{B}}=1 \\ {\mathrm{x}}_{\mathrm{B}}=\frac{1}{20} \end{gathered}$$

Step 4: Calculation for % mole for B:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{B}}=\frac{{\mathrm{x}}_{\mathrm{B}}\times \mathrm{P}{°}_{\mathrm{B}}}{\mathrm{P}}=\frac{{\displaystyle \frac{1}{20}}\times 80\mathrm{torr}}{99\mathrm{torr}}=\frac{4}{99} \\ {\mathrm{y}}_{\mathrm{B}}=\frac{4}{99} \\ \mathrm{So}, \\ \%{\mathrm{y}}_{\mathrm{B}}=\frac{4}{99}\times 100\approx 4\% \end{gathered}$$

Hence, the % mole for B is 4%.