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 phase and 0.2 in the liquid phase. The vapour pressure of pure liquid A is __________ atm. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

Find A's partial pressure from the vapour side, then use Raoult's law on the liquid side.

1. Vapour side. By Dalton's law, the partial pressure of A in the vapour is its vapour mole fraction times the total pressure: p_A = y_A P = 0.5 × 0.8 = 0.4 atm.
2. Liquid side. The vapour is in equilibrium with an ideal solution, so this same p_A also follows Raoult's law, using the liquid mole fraction x_A = 0.2.

   y_A P = x_A p_A° ⇒ p_A° = (y_A P)/x_A = 0.4/0.2

3. So. p_A° = 2 atm, so the answer is 2.

The common slip: 4. Dividing the total pressure by the liquid mole fraction (0.8/0.2) gives 4. Only A's partial pressure, 0.4 atm, belongs in Raoult's law for A.

The idea to keep. One partial pressure, two descriptions: y_A P from the vapour and x_A p_A° from the liquid. Set them equal, and keep each mole fraction on its own side.

Answer from the NTA final answer key. Solution by PrepWiser.