The vapour pressure of pure Benzene and Toluene are 900 torr and 600 torr respectively at 300K . A liquid solution is formed by mixing 1 mole each of benzene and toluene. If the pressure over the solution is reduced slowly under isothermal condition then mole fraction of benzene in the residue liquid, which has normal boiling point at 300K is:-

Question

The vapour pressure of pure Benzene and Toluene are 900 torr and 600 torr respectively at 300K . A liquid solution is formed by mixing 1 mole each of benzene and toluene. If the pressure over the solution is reduced slowly under isothermal condition then mole fraction of benzene in the residue liquid, which has normal boiling point at 300K is:- (A) 0.50 (B) 0.35 (C) 0.46 (D) 0.53

Tardigrade Admin · Accepted Answer

For residue solution at its normal boiling point: Ps=1atm=760 torr Ps=Xb× 900+X1× 600 760=900Xb+(1 - Xb)× 600=600+300Xb Xb=(160/300)=0.53

$0.50$

$0.35$

$0.46$

$0.53$

For residue solution at its normal boiling point:
$P_{s} = 1 a t m = 760$ torr
$P_{s} = X_{b} \times 900 + X_{1} \times 600$
$760 = 900 X_{b} + (1 - X_{b}) \times 600 = 600 + 300 X_{b}$
$X_{b} = \frac{160}{300} = 0.53$

0.50

0.35

0.46

0.53

For residue solution at its normal boiling point: Ps​=1atm=760 torr Ps​=Xb​×900+X1​×600 760=900Xb​+(1−Xb​)×600=600+300Xb​ Xb​=300160​=0.53

$0.50$

$0.35$

$0.46$

$0.53$

For residue solution at its normal boiling point:
$P_{s} = 1 a t m = 760$ torr
$P_{s} = X_{b} \times 900 + X_{1} \times 600$
$760 = 900 X_{b} + (1 - X_{b}) \times 600 = 600 + 300 X_{b}$
$X_{b} = \frac{160}{300} = 0.53$