Question

At ${100}^{\circ }$C, benzene & toluene have vapour pressure of 1375 & 558 Torr respectively. Assuming these two form an ideal binary solution, calculate the composition of lthe solution that boils at 1 atm & ${100}^{\circ }$C. What is the composition of vapour issuing at these conditions?

• A. $x_b=0.2472,Y_b=0.4473$
• B. $x_b=0.4473,Y_b=0.2472$
• C. $x_b=0.362,Y_b=0.321$
• D. None of these

Answer 1

The correct option is A $x_b=0.2472,Y_b=0.4473$

Let $X_B$ and $X_T$ be the mole fractions of benzene and toluene in solution.
$XT=1-XB$
$P_{total}=PB+PT$

$P_{total}$ is the total pressure of the mixture. It is $1$ atm or $760$ torr.

$P_B$ is the partial pressure of benzene. $P_T$ is partial pressure of toluene.
$P_B=X_B{P}_B^o=X_B\left(1375\right)$
$P_T=X_T{P}_T^o=(1-X_B)558$
$P_{total}=760=\left(1375\right)X_B+558(1-X_B)$
$760=\left(817\right)X_B+558$
$\left(817\right)X_B=202$
$X_B=0.2472$
$X_T=1-X_B=0.7528$
Let $Y_B$ and $Y_T$ be the mole fractions of benzene and toluene in the vapour phase.
$Y_B=\frac{P_B}{P_{total}}=\frac{1375\times 0.2472}{760}=0.4473$
$Y_T=1-0.4473=0.5527$