1‑propanol (p∘1=20.9 torr at 25 ∘c) and 2‑propanol (p∘2=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=0.220, calculate the composition of the vapor phase at 25 ∘c.

The composition of 1-propanol in vapor phase = 11.54 %

The composition of 2-propanol in vapor phase = 88.46 %

Explanation:

Mole fraction of components in liquid phase:

1-propanol =

2-propanol =

Partial pressure of 1-propanol =

Partial pressure of 2-propanol =

According to Raoults law:

Mole fraction of components in vapor phase:

1-propanol =

The composition of 1-propanol in vapor phase:

100 × 0.1154= 11.54 %

2-propanol =

The composition of 2-propanol in vapor phase:

100 × 0.8846 = 88.46 %

the answer is:  

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