Question

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.

Answer 1

Answer:

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 2

Answer: P= x×p°

2-propanol vapor pressure = (1-0.670)×45.2Torr= 14.916Torr

1-propanal vapor pressure = 0,670×20.9 Torr = 14.003 Torr

for calculating the composition : y = P/$P^{T}$

                                                       Total pressure = 14.196+ 14.003

                                                                                 = 28.919 Torr

(1-propanol) therefore = $\frac{14.003 Torr}{28.919Torr}$

                                     = 0.4841

(2-propanol)  = $\frac{14.816Torr}{28.919Torr}$ = 0.5158

Explanation: In an Ideal solution , sometimes mixtures obey Raoults law.

Using Raoults  law calculate the partial pressure of the component by using the mole fractions of each liquid.