Two liquids A and B have vapor pressures of 71.3 and 131 mmHg, respectively, at 25 degrees Celsius. What is the total vapor pressure of the ideal solution made up in the following concentrations?

Question

Two liquids A and B have vapor pressures of 71.3 and 131 mmHg, respectively, at 25 degrees Celsius. What is the total vapor pressure of the ideal solution made up in the following concentrations?

1.00 mole of A and 1.00 mole of B

2.00 moles of A and 5.00 moles of B

Answer 1

1 mol A + 1 mol B = 2 mols total.

X_A = nA/nTOTAL = 1/2
X_B = nB/nTOTAL = 1/2
P_A = X_A*P^o_A

P_B = X_B*P^o_B
Total P = P_A + P_B

Answer 2

To find the total vapor pressure of an ideal solution made up of two liquids, we can use Raoult's Law. Raoult's Law states that the partial pressure of each component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

The mole fraction of a component can be calculated by dividing the number of moles of that component by the total number of moles of all components.

Let's calculate the total vapor pressure for each case:

1. 1.00 mole of A and 1.00 mole of B:
To find the mole fraction of A and B, we divide the number of moles of each component by the total number of moles:
Mole fraction of A = (1.00 mole of A) / (1.00 mole of A + 1.00 mole of B) = 0.5
Mole fraction of B = (1.00 mole of B) / (1.00 mole of A + 1.00 mole of B) = 0.5

Now, we can use Raoult's Law to calculate the partial pressure of each component:
Partial pressure of A = vapor pressure of A * mole fraction of A = 71.3 mmHg * 0.5 = 35.65 mmHg
Partial pressure of B = vapor pressure of B * mole fraction of B = 131 mmHg * 0.5 = 65.5 mmHg

Finally, the total vapor pressure of the ideal solution is the sum of the partial pressures of each component:
Total vapor pressure = partial pressure of A + partial pressure of B = 35.65 mmHg + 65.5 mmHg = 101.15 mmHg

2. 2.00 moles of A and 5.00 moles of B:
Again, we calculate the mole fraction of each component:
Mole fraction of A = (2.00 moles of A) / (2.00 moles of A + 5.00 moles of B) = 0.286
Mole fraction of B = (5.00 moles of B) / (2.00 moles of A + 5.00 moles of B) = 0.714

Using Raoult's Law:
Partial pressure of A = vapor pressure of A * mole fraction of A = 71.3 mmHg * 0.286 = 20.42 mmHg
Partial pressure of B = vapor pressure of B * mole fraction of B = 131 mmHg * 0.714 = 93.534 mmHg

Total vapor pressure = partial pressure of A + partial pressure of B = 20.42 mmHg + 93.534 mmHg = 113.954 mmHg

So, the total vapor pressure of the ideal solution is 101.15 mmHg when there is 1.00 mole of A and 1.00 mole of B, and it is 113.954 mmHg when there are 2.00 moles of A and 5.00 moles of B.