Question

The vapour pressure of pure benzene at ${88}^{\circ }C$ is 600 mm Hg and that of toluene at the same temperature is 1240 mm Hg. Calculate the composition of a benzene-toluene mixture boiling at ${88}^{\circ }C$.

• A. $X_{benzene}=0.75,X_{toluene}=0.25$
• B. $X_{benzene}=0.25,X_{toluene}=0.75$
• C. $X_{benzene}=0.70,X_{toluene}=0.30$
• D. $X_{benzene}=0.30,X_{toluene}=0.70$

Answer 1

The correct option is A $X_{benzene}=0.75,X_{toluene}=0.25$

Since the B.P. is that temp. at which the vapour pressure of the liquid is equal to the atmospheric pressure i.e. 760 mm Hg.
$P=X_{benzene}{P}_{benzene}^{\circ }+X_{toulene}{P}_{toulene}^{\circ }$
$X_{benzene}+X_{toluene}=1$
$X_{toluene}=1-X_{benzene}$
$P=X_B{P}_B^{\circ }+(1-X_B)P_T^{\circ }=X_B{P}_B^{\circ }+P_T^{\circ }-P_T^{\circ }{X}_B$
$P=X_B(P_B^{\circ }-P_T^{\circ })+P_T^{\circ }$
$760=X_B(600-1240)+1240$
$X_B=\frac{760-1240}{600-1240}=\frac{-480}{-640}=0.75$
$X_T=1-X_B=1-0.75=0.25$