If a solution containing 4 mols of A and 6 mol 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 vapor pressure of pure B can be calculated using Raoult's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent times the vapor pressure of the pure solvent. Given that the total vapor pressure of the solution is 1 atm (760 mmHg) and the vapor pressure of A is 500 mmHg, we can calculate the partial pressure of A and then use the total pressure to find the vapor pressure of B. The mole fraction of A is 4/(4+6) = 0.4, and for B, it is 0.6. Since the vapor pressure of A contributes to part of the total pressure, we can find the contribution of B by subtracting the partial pressure of A from the total pressure. The calculation shows that the vapor pressure of B must be higher than the total pressure minus the contribution of A, leading to a value higher than the options provided, but since we are working within the constraints of the options, A (933 mmHg) is chosen as the correct answer by the process of elimination and understanding that the vapor pressure of B must be significantly contributing to reach the total pressure of 760 mmHg.

If a Solution Containing 4 Mols of a and 6