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.
If Vapours are removed and condensed in to liquid then what will be the ratio of mole fraction of benzene and toluene in first condensate :

• A. 2.8
• B. 1.5
• C. 3.5
• D. 4.5

Answer 1

The correct option is D 4.5

Given $P=179X_B+92$

For pure $C_6{H}_6,X_B=1$

$\therefore P_B^o=179+92=271mm$

For pure $C_7{H}_8,X_B=0$

$\therefore P_T^o=179\times 0+92=92mm$

Now, $P_M=P_B^o{X}_B+P_T^o{X}_T$

$=(271\times \frac{12}{12+8})+(92\times \frac{8}{12+8})$

$=199.4mm$

as, Moles of $C_6{H}_6=\frac{936}{78}=12$

Moles of $C_7{H}_8=\frac{736}{92}=8$

Now, mole fraction of $C_6{H}_6$ in vapour phase of initial mixture $X_B^1=\frac{162.6}{199.4}$

and that of toulene, $X_T^1=\frac{36.8}{199.4}$

$\therefore \frac{X_B^1}{X_T^1}=\frac{162.6}{36.8}=4.418$