A mixture of miscible liquids containing 6 moles of A and 4 moles of B is distilled 1 atm (760 mm Hg) of pressure. The initial distillate was analyzed and found to contain 20 mole percent A and 80 mole percent B.

a.) What are the partial vapor pressures of A and B in the initial mixture?
b.) What are the vapor pressures of pure A and B?

To determine the partial vapor pressures of A and B in the initial mixture, we need to consider their mole fractions and the total pressure of the system.

a.) Mole fraction is the ratio of the number of moles of a particular component to the total number of moles in the mixture. In this case, we have 6 moles of A and 4 moles of B, so the total number of moles in the mixture is 6 + 4 = 10 moles.

The mole fraction of A (X_A) can be calculated as:
X_A = (moles of A) / (total moles)
X_A = 6 / 10
X_A = 0.6

Similarly, the mole fraction of B (X_B) can be calculated as:
X_B = (moles of B) / (total moles)
X_B = 4 / 10
X_B = 0.4

The partial vapor pressure of A (P_A) can be calculated using Raoult's law, which states that the partial vapor pressure of a component in an ideal mixture is equal to its mole fraction multiplied by the vapor pressure of the pure component. Assuming ideal behavior, we can use the mole fractions and the given total pressure of 1 atm (760 mm Hg) to calculate P_A:
P_A = X_A * P_total
P_A = 0.6 * 1 atm (760 mm Hg)
P_A = 0.6 atm (456 mm Hg)

Similarly, the partial vapor pressure of B (P_B) can be calculated as:
P_B = X_B * P_total
P_B = 0.4 * 1 atm (760 mm Hg)
P_B = 0.4 atm (304 mm Hg)

b.) To find the vapor pressures of pure A and B, we need to determine their respective mole fractions in the distillate.

Given that the distillate contains 20 mole percent A and 80 mole percent B, we can calculate the mole fraction of A (X_A_distillate) in the distillate as:
X_A_distillate = 20 / 100
X_A_distillate = 0.2

Similarly, the mole fraction of B (X_B_distillate) in the distillate can be calculated as:
X_B_distillate = 80 / 100
X_B_distillate = 0.8

Now, we can apply Raoult's law to calculate the vapor pressures of pure A and B.

For pure A:
P_A_pure = X_A_distillate * P_total
P_A_pure = 0.2 * 1 atm (760 mm Hg)
P_A_pure = 0.2 atm (152 mm Hg)

For pure B:
P_B_pure = X_B_distillate * P_total
P_B_pure = 0.8 * 1 atm (760 mm Hg)
P_B_pure = 0.8 atm (608 mm Hg)

Therefore, the partial vapor pressure of A in the initial mixture is 0.6 atm (456 mm Hg), the partial vapor pressure of B in the initial mixture is 0.4 atm (304 mm Hg), the vapor pressure of pure A is 0.2 atm (152 mm Hg), and the vapor pressure of pure B is 0.8 atm (608 mm Hg).