At 363 K, the vapour pressure of A is 21 kPa and that of B is 18 kPa. One mole of A and 2 moles of B are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is ___________ kPa. (Round off to the Nearest Integer).

Solution

Chemistry ma’am · 3 steps

Turn the moles into mole fractions, then let each liquid add its share of the pressure.

1. Mole fractions. There are 1 + 2 = 3 moles in total. So x_A = 1/3 and x_B = 2/3.
2. Raoult's law. In an ideal solution, each component's partial pressure is its mole fraction times its pure vapour pressure. The total pressure is the sum of the two partial pressures.

   P = x_A P_A° + x_B P_B° = 1/3(21) + 2/3(18) = 7 + 12

3. So. The vapour pressure of the mixture is 19 kPa, so the answer is 19.

The common slip: 20. Giving A the larger mole fraction 2/3 makes the total 14 + 6 = 20 kPa, not 19.

The idea to keep. For an ideal mixture, the total vapour pressure is a mole-fraction-weighted average of the pure vapour pressures. Always pair each mole fraction with its own liquid's P°.

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