Question

A liquid mixture containing 26 g $C_6{H}_6$ and 46 g $C_7{H}_8$ at ${50}^{o}C$ has a vapour pressure of 163.75 mm of Hg. When another 52 g of $C_6{H}_6$ is added, V.P. of mixture is increased to 211.57 mm of Hg. Find the values of A (divide by 30) if vapour pressure of pure $C_6{H}_6$ and toluene are represented in pure state by $P=A+B\cdot X_T(X_T$ is mole fraction of toluene).

Answer 1

The molar masses of benzene and toluene are $78$ g/mol and $92$ g/mol respectively.

$26$ g benzene $=\frac{26}{78}=0.333$ moles

$46$ g toluene $=\frac{46}{92}=0.5$ moles

Mole fraction of benzene $=X_B=\frac{0.333}{0.333+0.5}=0.4$

Mole fraction of toluene $X_T=1-0.4=0.6$

Total pressure $P=P_B^o{X}_B+P_T^o{X}_T$

$163.75=P_B^o\times 0.4+P_T^o\times 0.6$......(1)

Another $52$ g of benzene are added

Number of moles of benzene $=\frac{52+26}{78}=1$

Mole fraction of benzene $X_B=\frac{1}{1+0.5}=0.667$

Mole fraction of toluene $X_T=1-0.667=0.334$

Total pressure $P=P^{o}BXB+P_T^o{X}_T$

$211.57=P_B^o\times 0.667+P_T^o\times 0.334$......(2)

Solving the simultaneous equations (1) and (2) we get

$P_B^0=271.35mm,P_T^0=92.02mm$

$P=A+B{X}_T=P_B^o+(P_T^o-P_B^o)XT$

$A=P_B^o=271.35mm$

$B=(P_T^o-P_B^o)=(92.02-271.35)=-179.33mm$