Respuesta :
Answer:
a)Pheptane = 24.3 torr
Poctane = 5.12 torr
b)Ptotal vapor = 29.42 torr
c) 81 % heptane
19 % octane
d) See explanation below
Explanation:
The partial pressure is given by Raoult´s law as:
Pa = Xa Pºa where Pa = partial pressure of component A
Xa = mole fraction of A
Pºa = vapor pressure of pure A
For a binary solution what we have to do is compute the partial vapor pressure of each component and then add them together to get total vapor pressure.
In order to calculate the composition of the vapor in part b), we will first calculate the mole fraction of each component in the vapor which is given by the relationship:
Xa = Pa/Pt where Xa = mol fraction of in the vapor
Pa = partial pressure of A as calculated above
Pt = total vapor pressure
Once we have mole fractions we can calculate the masses of the components for part c)
a)
MW heptane = 100.21 g/mol
MW octane = 114.23 g/mol
mol heptane = 50.0 g / 100.21 g/mol = 0.50 mol
mol octane = 50.0 g/ 114.23 g/mol = 0.44 mol
mol total = 0.94 ⇒ Xa= 0.50/0.94 = 0.53 and
Xb= 0.44/0.94 = 0.47
Pheptane = 0.53 x 45.8 torr = 24.3 torr
Poctane = 0.47 x 10.9 torr = 5.12 torr
b) Ptotal = 24.3 torr +5.12 torr = 29.42 torr
c) We will call Y the mole fraction in the vapor to differentiate it from the mole fraction in solution
Y heptane (in the vapor) = 24.3 torr/ 29.42 torr = 0.83
Y octane (in the vapor) = 5.12 torr/ 29.42 torr = 0.17
d) To solve this part we will assume that since the molecular weights are similar then having a mole fraction for heptane of 0.82, we could say that for every mole of mixture we have 0.82 mol heptane and 0.17 mol octane and then we can calculate the masses:
0.82 mol x 100.21 g/mol = 82.2 g
0.17 mol x 114.23 g/mol = 19.4 g
total mass = 101.6
% heptane = 82.2 g/101.6g x 100 = 81 %
% octane = 19 %
There is another way to do this more exactly by calculating the average molecular weight of the mixture:
average MW = 0.83 (100.21 g/mol) + 0.17 ( 114.23 g/mol ) = 102. 6 g/mol
and then having a mol fraction of 0.83 means in 1 mol of mixture we have 0.83 mol heptane and 0.17 mol octane then the masses are:
mass heptane = 0.83 x 100.21 g/mol = 83.2 g
mass octane = 0.17 x 114.23 g/mol = 19.4 g
mass of mixture = 1 mol x MW mixture = 1 mol x 102.6 g/mol 102.6 g
% heptane = (83.2 g/ 102.6 g ) x 100 g = 81 %
% octane = 100 - 81 = 19 %
d)The composition of the vapor is different from the composition of the solution because the vapor is going to be richer in the more volatile compound in the solution which in this case is heptane ( 45.8 vs 10.9 torr).
The vapor will contain more of the volatile substance than the solution.
What is vapor pressure?
This is the pressure exerted on the container when a substance is converted to vapor phase.
Molar mass of heptane= 100.21 g/mol
Molar mass of octane= 114.23 g/mol
Number of moles of heptane = 50.0 g / 100.21 g/mol = 0.50 mol
Number of moles of octane= 50.0 g/ 114.23 g/mol = 0.44 mol
Total number of moles = 0.94
Let the mole fraction of heptane be Xa
Xa= 0.50/0.94 = 0.53
Let the mole fraction of octane be Xb
Xb= 0.44/0.94 = 0.47
Partial pressure of heptane = 0.53 x 45.8 torr = 24.3 torr
Partial pressure of octane= 0.47 x 10.9 torr = 5.12 torr
b)Total pressure of the mixture = 24.3 torr +5.12 torr = 29.42 torr
c)Let the mole fraction of the vapor be Y
Y (heptane) = 24.3 torr/ 29.42 torr = 0.83
Y (octane) = 5.12 torr/ 29.42 torr = 0.17
d)
Mass of heptane =0.82 mol x 100.21 g/mol = 82.2 g
Mass of octane =0.17 mol x 114.23 g/mol = 19.4 g
Total mass present = 101.6
Percentage of heptane = 82.2 g/101.6g x 100 = 81 %
Percentage of octane = (100 - 81 %) = 19 %
d)The vapor will contain more of the volatile substance than the solution.
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