Question

if the solution containing 4 mols of A and 6 mols of B boils at 85 degree and the vapor pressure of pure A at 85 is 500mmHg, what is the vapor pressure of pure B at this temperature?

Answer 1

The vapor pressure of pure B at 85 degrees Celsius is approximately 933.33 mmHg.

To find the vapor pressure of pure B at 85 degrees Celsius when you have a solution containing 4 moles of A and 6 moles of B that boils at 85 degrees Celsius, you can use Raoult's law. Raoult's law describes the vapor pressure of an ideal solution as a function of the mole fractions of the components and the vapor pressures of the pure components. It can be expressed as:

`P_total = x_A * P_A + x_B * P_B`

Where:

- Ptotal is the vapor pressure of the solution.

- xA and xB are the mole fractions of A and B in the solution, respectively.

- PA and PB are the vapor pressures of pure A and pure B at the given temperature.

In this case, you are given that the solution contains 4 moles of A and 6 moles of B. You can calculate the mole fractions of A and B as follows:

Mole fraction of A (xA) = Moles of A / Total moles in the solution

`x_A = 4 / (4 + 6) = 4/10 = 0.4`

Mole fraction of B (xB) = Moles of B / Total moles in the solution

`x_B = 6 / (4 + 6) = 6/10 = 0.6`

You are also given that the solution boils at 85 degrees Celsius, and the vapor pressure of pure A at this temperature is 500 mmHg. Therefore, you can use Raoult's law to find the vapor pressure of pure B (P_B) at 85 degrees Celsius:

`P_total = x_A * P_A + x_B * P_B`

Ptotal = 1 atm (since it's boiling, it's at its boiling point, and we can assume it's at 1 atm)

Now, you need to rearrange the equation to solve for P_B:

`P_B = (P_total - x_A * P_A) / x_B`

Convert the pressure to mmHg (1 atm = 760 mmHg):

`P_total = 1 atm * 760 mmHg/atm = 760 mmHg`

Now, plug in the values:

PB = (760 mmHg - 0.4 * 500 mmHg) / 0.6

PB = (760 mmHg - 200 mmHg) / 0.6

PB = 560 mmHg / 0.6

PB ≈ 933.33 mmHg