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: 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/[tex]P^{T}[/tex]
Total pressure = 14.196+ 14.003
= 28.919 Torr
(1-propanol) therefore = [tex]\frac{14.003 Torr}{28.919Torr}[/tex]
= 0.4841
(2-propanol) = [tex]\frac{14.816Torr}{28.919Torr}[/tex] = 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.
