Question

The vapour pressure of pure benzene and toluene are 160mm and 60mm of hg respectively.the mole fraction of benzene in vapogur phase in contact with equimolar solution of benzene and toluene is

Answer 1

Answer: The mole fraction of benzene in vapor phase is 0.73

Explanation: According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_A=x_Ap_A^0$ and $p_B=x_BP_B^0$

where, x = mole fraction

$p^0$ = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_A+p_B$

$p_{total}=x_Ap_A^0+x_BP_B^0$

given $x_{benzene}=0.5$, $x_{toluene}=0.5$  (as they have equal moles)

$p_{benzene}^0=160mmHg$

$p_{toluene}^0=60mmHg$

$p_{total}=0.5\times160+0.5\times 60=110mmHg$

The mole fraction of benzene in vapour phase is given by:

$y_{benzene}=\frac{p_{benzene}}{P_{total}}$

$p_{benzene}=x_{benzene}\times p_{benzene}^0=0.5\times 160=80mmHg$  

$y_{benzene}=\frac{80mmHg}{110mmHg}=0.73$