The vapour pressure of pure liquids A and B are 450 and 750 mm Hg respectively, at 350K. Find out the composition ofthe liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

To solve this problem, we will use Raoult's Law, which states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Let's denote the mole fraction of A and B in the liquid phase as xA and xB respectively.

Given:

• The vapour pressure of pure A (p°A) = 450 mm Hg
• The vapour pressure of pure B (p°B) = 750 mm Hg
• The total vapour pressure above the solution (p) = 600 mm Hg

According to Raoult's Law, the total pressure is the sum of the partial pressures of A and B:

p = xAp°A + xBp°B

Since xA and xB are mole fractions, xA + xB = 1. Therefore, we can express xB as 1 - xA.

Substituting xB in the equation, we get:

p = xA*p°A + (1 - xA)*p°B

Rearranging the equation to solve for xA:

xA = (p - p°B) / (p°A - p°B)

Substituting the given values:

xA = (600 - 750) / (450 - 750) = 0.6

Therefore, the mole fraction of B in the liquid phase, xB = 1 - xA = 1 - 0.6 = 0.4

Now, let's find the composition of the vapour phase. The partial pressure of A (pA) and B (pB) can be calculated as:

pA = xAp°A = 0.6450 = 270 mm Hg pB = xBp°B = 0.4750 = 300 mm Hg

The mole fraction of A and B in the vapour phase (yA and yB) can be calculated as:

yA = pA / p = 270 / 600 = 0.45 yB = pB / p = 300 / 600 = 0.55

So, the composition of the liquid mixture is 60% A and 40% B, and the composition of the vapour phase is 45% A and 55% B.