Question

Liquids A and B form an ideal solution in the entire compositon range. At 350 K, the vapor pressures of pure A and pure B are $7\times {10}^3$ Pa and $12\times {10}^3$ Pa, respectively. The composition of the vapor in equilibrium with a solution containing 40 mole percent of A at this temperature is:

• A. $X_A=0.28;X_B=0.72$
• B. $X_A=0.76;X_B=0.24$
• C. $X_A=0.37;X_B=0.63$
• D. $X_A=0.4;X_B=0.6$

Answer 1

The correct option is A $X_A=0.28;X_B=0.72$

Given:
$\text{Vapour pressure of pure A}\left(P_A^0\right)=7\times {10}^3\text{Pa}$
$\text{Vapour pressure of pure B}\left(P_B^0\right)=12\times {10}^3\text{Pa}$
$\text{Moles of}A=40\%=0.4$
$\text{Moles of}B=60\%=0.6$
$\therefore {\chi }_A=\frac{0.4}{0.4+0.6}=0.4$,
${\chi }_B=0.6$
$\text{Composition of vapour}\left(X_A\right)=\frac{P_A}{P_{Total}}=\frac{P_A^0{\chi }_A}{P_A^0{\chi }_A+P_B^0{\chi }_B}$
$=\frac{7\times {10}^3\times 0.4}{7\times {10}^3\times 0.4+12\times {10}^3\times 0.6}=\frac{2.8}{10}=0.28$
$\therefore X_B=1-0.28=0.72$