Question

A mixture contains 1 mole of volatile liquid a (=100 mm hg) and 3 moles of volatille liquid b (= 80 mm hg). If solution behaves ideally, the total vapour pressure of the distillate is

Answer 1

The total vapour pressure of the distillate is 85 mm of Hg

Explanation :

Total number of moles = 1 + 3 = 4

Mole fraction of A, Xa = 1/4 = 0.25

Mole fraction of B, Xb = 3/4 = 0.75

given partial pressure of A = 100 mm

partial pressure of B = 80 mm

we know that total vapour pressure of a distillate

P = PaXa + PbXb

= 100 x 0.25 + 80 x 0.75

= 25 + 60

=> P = 85 mm of Hg

Hence the total vapour pressure of the distillate is 85 mm of Hg

Answer 2

Answer:

85.88

Explanation:

- pA=p∘AxA=100×14 = 25 mm Hg

pB=p∘BxB=80×34 = 60 mmHg

Mole fraction of A in vapour

x′A=pApA+pB=25÷85

Mole fraction of B in vapour

x′B=1−25÷85=60÷85

V.P. of distillate

= x′Ap∘A+x′Bp∘B

= 25÷85×100+60÷85×80

= 1÷85(2500+4800)

= 7300÷85 = 85.88 = 86 mm Hg.