If a solution containing 4 mols of A and 6 mols of B boils at 85°C and 1 atm of pressure, and the vapor pressure of pure A at 85°C is 500 mmHg, what is the vapor pressure of pure B at this temperature?
A) 933 mmHg
B) 336 mmHg
C) 260 mmHg
D) 576 mmHg
E) 960 mmHg

Question

Quizplus · Accepted Answer

The total vapor pressure of the solution is 1 atm (760 mmHg). Using Raoult's Law, the partial pressure of A is (4/10) * 500 mmHg = 200 mmHg. Therefore, the partial pressure of B is 760 mmHg - 200 mmHg = 560 mmHg. The mole fraction of B is 6/10, so the vapor pressure of pure B is 560 mmHg / (6/10) = 933 mmHg.

If a Solution Containing 4 Mols of a and 6