Question

Liquid $A$ and $B$ form ideal solution at temperature $T$. Mole fraction of $A$ in liquid and vapor phase are $0.4$ and $4/3$ respectively when total pressure is $130$ torr. The vapor pressure (in torr) of $A$ and $B$ in pure state at temperature $T$ are respectively.

• A. $100,200$
• B. $200,100$
• C. $100,150$
• D. $150,100$

Answer 1

The correct option is C $100,150$

$\text{Given A and}B\text{are ideal solution at temp T}$

$\begin{array}{c}{x}_A\text{in}liquid\text{phase}=0.4\\ x_A\text{in vapour phase}=\frac{4}{13}\\ \text{Total pressure}{P}_T=130\\ P_A\&P_B=?\end{array}$

$P_T=P_A^0{X}_A+P_B^0{X}_B$

$\begin{array}{c}{P}_A=P_A^0\cdot X_A=\text{(Gas phase)}\frac{4}{13}\times 130=40\end{array}$

$P_A^0=\frac{P_A}{X_A}\text{(liquid phase)}=\frac{40}{0\cdot 4}=100$

$P_B=P_B^0\times X_B=\frac{9}{13}\times 130=90\text{in gas phase}$

$P_B^0=\frac{P_B}{X_B}=\frac{90}{0.6}=150\text{atm liquid phase}$

$P_A^0=100,P_B^0=150$

Option C is correct answer