Question

Vapour Pressure of a mixture of benzene and toluene is given by $P=179X_B+92$, Where $X_B$ is mole fraction of benzene.
This condensed liquid again brought to the same temperature then what will be the mole fraction of benzene in vapour phase :

• A. 0.007
• B. 0.93
• C. 0.65
• D. 4.5

Answer 1

The correct option is B 0.93

Mole fraction of $C_6{H}_6$ in vapour phase $=X_B^1$ of initial mixture

$X_B^1=\frac{P_B^1}{P_M}=\frac{162.6}{199.4}=0.815$

then mole fraction of $C_7{H}_8=0.185$

(mole fraction of $C_6{H}_6$ in initial mixture) $=$ (mole fraction of $C_6{H}_6$ in liquid phase on $II$ mixture $X_B^1$)

$P_M=P_B^1+P_T^1$

now, $P_M=P_B^0{X}_B^1+P_T^0{X}_T^1$

$=[\left(271\right)\left(0.815\right)+\left(92\right)\left(0.185\right)]$ mm

$=237.885$ mm

new mole fraction of $C_6{H}_6=\frac{220.865}{237.885}=0.928\cong 0.93$

$\therefore$ new mole fraction of Benzene in vapour phase is $0.93$.