Question

5 mole of liquid A and 5 mole of liquid B form an ideal solution at 25oC given  = 300 mm Hg and  = 500 mm Find composition of first bubble of vapour​

Answer 1

Answer:

500

Explanation:

500-300=200

200+300=500

Answer 2

WE ARE GIVEN,

vapour pressure of A= 300mmHg

vapour pressure of B= 500mmHg

number of moles of A=5

number of moles of B=5

Mole fraction of A(xₐ)= $\frac{no of moles of A}{total moles}$ = $\frac{5}{10}$ = 0.5                            

Mole fraction of B(Xᵇ)=  1- xₐ = 1- 0.5= 0.5

By applying Raoult's law  

PA= PA*Xa = 0.5×300mmHg = 150mmHg

PB= PB*Xb= 0.5×500mmHg =250mmHg

Mole fraction of A at vapor phase = $\frac{Pa}{Pa+Pb}$ =$\frac{150}{150+250}= \frac{150}{400}= 0.375$

HENCE THE COMPOSITION OF THE FIRST BUBBLE IS 0.375.