Question

1-propanol (P1° = 20.9 Torr at 25 °C) and 2-propanol (P2° = 45.2 Torr at 25 °C) form ideal solutions in all proportions. Let x1 and x2 represent the mole fractions of 1-propanol and 2-propanol in a liquid mixture, respectively, and y1 and y2 represent the mole fractions of each in the vapor phase. For a solution of these liquids with x1 = 0.450, calculate the composition of the vapor phase at 25 °C.

y1= y2=

Answer 1

y1 = 0.3162

y2 = 0.6838

Step-by-step explanation:

ok let us begin,

first we would be defining the parameters;

at 25°C;

1-propanol P1° = 20.90 Torr

2-propanol P2° = 45.2 Torr

From Raoults law:

P(1-propanol) = P⁰ × X(1-propanol)

P(1-propanol) = 20.9 torr × 0.45 = 9.405

P(1-propanol) = 9.405 torr

Also P(2-propanol) = P⁰ × X(2-propanol)

P(2-propanol) = 45.2 torr × 0.45

P(2-propanol) = 20.34 torr

but the total pressure = sum of individual pressures

total pressure = 9.405 + 20.34

total pressure = 29.745 torr

given that y1 and y2 represent the mole fraction of each in the vapor phase

y1 = P1 / total pressure

y1 = 9.405/29.745

y1 = 0.3162

Since y1 + y2 = 1

y2 = 1 - y1

∴ y2 = 1 - 0.3162

y2 = 0.6838

cheers, i hope this helps.