To solve the problem, we will break it down into the required parts step by step.
### Given Data:
- Vapor pressure of pure liquid A, \( P^0_A = 300 \, \text{torr} \)
- Vapor pressure of pure liquid B, \( P^0_B = 800 \, \text{torr} \)
- Mole fraction of A in vapor phase, \( y_A = 0.25 \)
- Mole fraction of B in vapor phase, \( y_B = 1 - y_A = 0.75 \)
### (a) Composition of the first drop of the condensate (Mole fraction of A in liquid phase, \( x_A \)):
Using Raoult's Law, the total pressure \( P_{total} \) can be calculated as follows:
\[
\frac{1}{P_{total}} = \frac{y_A}{P^0_A} + \frac{y_B}{P^0_B}
\]
Substituting the values:
\[
\frac{1}{P_{total}} = \frac{0.25}{300} + \frac{0.75}{800}
\]
Calculating the right-hand side:
\[
\frac{1}{P_{total}} = \frac{0.25}{300} + \frac{0.75}{800} = \frac{0.0008333}{1} + \frac{0.0009375}{1} = 0.0017708
\]
Now, calculating \( P_{total} \):
\[
P_{total} = \frac{1}{0.0017708} \approx 564.7 \, \text{torr}
\]
Now, we can use the total pressure to find the mole fraction of A in the liquid phase \( x_A \):
Using the equation:
\[
P_{total} = x_A P^0_A + x_B P^0_B
\]
Where \( x_B = 1 - x_A \):
\[
564.7 = x_A \cdot 300 + (1 - x_A) \cdot 800
\]
Expanding and rearranging gives:
\[
564.7 = 300x_A + 800 - 800x_A
\]
\[
564.7 = 800 - 500x_A
\]
\[
500x_A = 800 - 564.7
\]
\[
500x_A = 235.3
\]
\[
x_A = \frac{235.3}{500} \approx 0.4706
\]
Thus, the composition of the first drop of the condensate is:
\[
\boxed{0.47}
\]
### (b) Total pressure when this drop is formed:
From the calculation above, we have already found:
\[
\boxed{564.7 \, \text{torr}}
\]
### (c) Composition of the solution whose normal boiling point is T:
At the normal boiling point, the total pressure is equal to atmospheric pressure (760 torr):
Using the equation:
\[
760 = x_A P^0_A + x_B P^0_B
\]
Substituting \( x_B = 1 - x_A \):
\[
760 = x_A \cdot 300 + (1 - x_A) \cdot 800
\]
Expanding and rearranging gives:
\[
760 = 300x_A + 800 - 800x_A
\]
\[
760 = 800 - 500x_A
\]
\[
500x_A = 800 - 760
\]
\[
500x_A = 40
\]
\[
x_A = \frac{40}{500} = 0.08
\]
Thus, the composition of the solution at normal boiling point is:
\[
\boxed{0.08}
\]
### (d) Pressure when only the last bubble of vapor remains:
When only the last bubble of vapor remains, the composition of the vapor is the same as the composition of the liquid. Therefore:
Using the previously calculated mole fractions:
\[
x_A = 0.25 \quad \text{and} \quad x_B = 0.75
\]
Calculating the total pressure:
\[
P_{total} = x_A P^0_A + x_B P^0_B
\]
\[
P_{total} = 0.25 \cdot 300 + 0.75 \cdot 800
\]
\[
P_{total} = 75 + 600 = 675 \, \text{torr}
\]
Thus, the pressure when only the last bubble of vapor remains is:
\[
\boxed{675 \, \text{torr}}
\]
### (e) Composition of the last bubble:
The composition of the last bubble is the same as the composition of the liquid at that point:
Using the previously calculated values:
\[
x_A = 0.11 \quad \text{and} \quad x_B = 1 - x_A = 0.89
\]
Thus, the composition of the last bubble is:
\[
\boxed{0.11}
\]
