Question

The vapour pressure of a pure liquid A is 40 mm of Hg at 310 K. The vapour pressure of this liquid in a solution with liquid B is 32 mm of Hg. The mole fraction of A in the solution, if it obey's Raoult's law, is?

• A. 0.8
• B. 0.5
• C. 0.2
• D. 0.4

Answer 1

The correct option is A 0.8

P${}_A$ = 40mm of Hg

P${}_T$ =32 mm of Hg

By Roult's Law, P${}_T=P_A\times X_A$

32= X${}_A\times$ 40

X${}_A=\frac{32}{40}$

$X_A$=0.8

Therefore, mole fraction of A in the solution is 0.8