Question

Two liquids A and B on mixing form ideal solution. their vapour pressures in pure state are 200 and 100 mm respectively. what will be the mole frction of B in the vapour phase in equillibrium with an equimolar solution of two?

Answer 1

Let the moles of A=B = m
Mole fraction of B = m/m+m =m/2m
According to Raoults law
P=P°x
P(B)=P°(B)x(B)
x(B) = m/2m
P°(B) = 100
P(B) = 100*m/2m = 50mm
Similarly
P(A) = 100mm
Y(B) = P(B)/P(A)+P(B)
= 50/100+50 = 1/3

Answer 2

Answer: The mole fraction of B in vapor state is 0.33.

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 $x_{A}=0.5$, $x_{B}=0.5$

$p_{A}^0=200 mmHg$

$p_{B}^0=100 mmHg$

$p_{total}=0.5\times200+0.5\times 100=150mmHg$

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

$y_{B}=\frac{p_B}{P_{total}}$

$p_{B}=x_{B}\times p_{B}^0=0.5\times 100=50mmHg$  

$y_{B}=\frac{50mmHg}{150mmHg}=0.33$