Question

An ideal solution of two volatile liquid $A$ and $B$ has a vapour pressure of $402.5mmHg$, the mole of fraction of $A$ in vapour & liquid state being $0.35$ & $0.65$ respectively. What are the vapour pressure of the two liquid at this temperature.

• A. $P_A^{\circ }=216.7mmHg,P_B^{\circ }=747.5mmHg$
• B. $P_A^{\circ }=747.5mmHg,P_B^{\circ }=216.7mmHg$
• C. $P_A^{\circ }=528.7mmHg,P_B^{\circ }=432.5mmHg$
• D. $P_A^{\circ }=432.5mmHg,P_B^{\circ }=528.7mmHg$

Answer 1

The correct option is A $P_A^{\circ }=216.7mmHg,P_B^{\circ }=747.5mmHg$

$X_A=0.65X_B=1-XA=1-0.65=0.35Y_A=0.35Y_B=1-Y_A=1-0.35=0.65$
$P_T=P_A^0{X}_A+P_B^0{X}_{B}402.5=P_A^0\times 0.65+P_B^0\times 0.35$ ......(1)
$P_A=P_A^0{X}_A=P_A^0\times 0.65$.......(2)
$P_A=P_T{Y}_A=402.5\times 0.35=140.875$......(3)
Substitute (3) in (2)
$140.875=P_A^0\times 0.65$
$P_A^0=216.7$ mm Hg.....(4)
Substitute (4) in (1)
$402.5=216.7\times 0.65+P_B^0\times 0.35$
$P_B^0=747.5$ mm Hg