A gaseous mixture of two substances A and B, under a total pressure of 0.8 atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is 0.5 in the vapour phae and 0.2 in the liquid phase. The vapour pressure of pure liquid A is ___________________ atm.

(Nearest integer)

We are given the following data in the problem:

• mole fraction of substance A in the liquid phase, X_A = 0.2
• mole fraction of substance A in the vapour phase, Y_A = 0.5
• total pressure of the system, P_T = 0.8 atm

We are asked to calculate the vapour pressure of pure liquid A, denoted as P_A°.

Step-by-step Calculation:

1. We know that the partial pressure of component A in the vapour phase, P_A, can be calculated using the equation:  P_A = Y_A × P_T 

    Substituting the values:  P_A = 0.5 × 0.8 = 0.4 atm.

2. Next, we use Raoult’s Law, which states that the partial pressure of component A in the liquid phase is related to the vapour pressure of pure liquid A (P_A°) and the mole fraction of A in the liquid phase (X_A) by the equation: 
   P_A = X_A × P_A°
3. We already know that P_A = 0.4 atm and X_A = 0.2. So, substituting these values into the equation: 
   0.4 = 0.2 × P_A°
4. Solving for P_A°:  P_A° = 0.4 / 0.2 = 2 atm.

Final Answer:

The vapour pressure of pure liquid A is 2 atm.