Question

A solution contains a mixture of two volatile substances A and B. The mole fraction of substance A is 0.39. At 32∘C the vapor pressure of pure A is 95 mmHg and the vapor pressure of pure B is 141 mmHg. What is the total vapor pressure of the solution at this temperature?

Answer 1

The total vapor pressure of the solution at 32°C is 123.06 mmHg.

Step-by-step explanation:

Dalton's law of partial pressures states that the total pressure is equal to the sum of the partial pressures. In this case, we need to use Raoult's law to calculate the vapor pressure of the solution. Raoult's law states that the vapor pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure in the pure state.

Given that the mole fraction of substance A is 0.39 and its vapor pressure is 95 mmHg, we can calculate the partial pressure of substance A in the solution as follows:

PA = XAPA = 0.39 * 95 mmHg = 37.05 mmHg

Similarly, the partial pressure of substance B can be calculated using the mole fraction of B (1 - 0.39 = 0.61) and its vapor pressure of 141 mmHg:

PB = XB * P = 0.61 * 141 mmHg = 86.01 mmHg

The total vapor pressure of the solution at 32°C is the sum of the partial pressures of substances A and B:

Ptot = PA + PB = 37.05 mmHg + 86.01 mmHg = 123.06 mmHg