Question

The vapour pressure of a pure liquid a is 40 mm hg at 310k. the vapour pressure of this liquid in a solution with liquid b is 32 mm hg. mol fraction of a in the solution if it obeys raoult's law is:

Answer 1

as per Raoult's law pressure of pure liquid in a solution(pi) is equals to mole fraction(X) of it × pressure of it in solution form (p)
therefore
p=X.pi

32=X.40
X=32/40
X=0.8

Answer 2

According to Roult’s law,
Where,
P is the pressure of solution
P0 is the pressure of pure solvent
And xsolv is the mole fraction of the solvent
Substituting the given values in the equation we get,
32 = xsolv x 40
xsolv = 32/40
  = 0.8
Hence, the mole fraction of liquid A in the solution = 0.8