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 x 1 and x 2 represent the mole fractions of 1‑propanol and 2‑propanol in a liquid mixture, respectively, and y 1 and y 2 represent the mole fractions of each in the vapor phase. For a solution of these liquids with x 1 = 0.670 , calculate the composition of the vapor phase at 25 ∘ C.

y₁ = 0.48

y₂ = 0.52

Explanation:

The method to solve this question is to use Raoult´s law for ideal solutions, which tell us that the vapor pressure of a component A in solution is equal to:

Pa = Xa Pºa

where Pa is the partial pressure of a, xa is its mole fraction, and Pºa is the vapor pressure of pure A.

From here it follows that for a binary solution the total pressure is the sum of the partial pressures of each component.

With vthis in mind we are ready to calculate and solve our question:

P1 = x₁Pº₁ = 0.670 x 20.9Torr = 14.00 torr

P₂ = x₂Pº₂ = (1-0.670) x 45.2 torr =  0.33 x 45.2 torr = 14.91 torr

Ptotal = 14.00 torr + 14.91 torr = 28.91 torr

The composition of the vapor will be given by:

y₁ = Py₁ / Ptotal = 14.00 torr/ 28.91 torr = 0.48

y₂ = 1 - y₁ = 1 - 0.48 = 0.52

answer: c.

explanation: