Question

Vapour pressure of C6H6 and C7H8 mixture at 50°C is given by P (mm Hg) = 179XB + 92, where XB is the mole fraction of C6H6 . A solution is prepared by mixing 936 g benzene and 736 g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of 50°C, what would be mole fraction of C6H6 in the vapour state?

Answer 1

Step-by-step explanation:

We know the equation

P= PoA +Xb (PoB -PoA)

Po Tol = 92

Po Ben = 179+92 = 271

n ben = 936/78

n tol = 736/92

Solving we get

X ben = 0.92

Answer 2

Answer:

Mole fraction of $C_{6} H_{6}$ in the vapour state  is 0.9286

Step-by-step explanation:

TO FIND:

       Mole fraction of $C_{6} H_{6}$ in the vapour state

           $P=179 X_{B} +92$

      Vapour pressure of C6H6 and C7H8 mixture at 50°C is given by $P (mm Hg) = 179XB + 92$,  where XB is the mole fraction of C6H6.

 

A solution is prepared by mixing 936 g benzene and 736 g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of 50°C.

$P_{B}$ =271, $P_{T}$ =92

$n_{B}=\frac{936}{78}\\=12$

$n_{T} =\frac{736}{92} \\=8$

$X_{B} =\frac{12}{20} \\0.6$

$X_{T}=0.4$

$P_{T} =271*0.6+92*0.4\\=199.4$

$Y_{B} =\frac{271*0.6}{199.4}\\=0.815$

$Y_{T} =0.185$

On additional buildup

$X_{B} =0.185,X_{T} =0.185$,

$P_{T} = 271*0.815+92*0.185\\=237.844$

$Y_{B} =\frac{271*0.815}{237.844} \\=0.9286$

Mole fraction of $C_{6} H_{6}$ in the vapour state  is 0.9286

The project code is #SPJ3