A solution contains 43.0 g of heptane (C7H16) and 42.5 g of octane (C8H18) at 25
ID: 480568 • Letter: A
Question
A solution contains 43.0 g of heptane (C7H16) and 42.5 g of octane (C8H18) at 25 C . The vapor pressures of pure heptane and pure octane at 25 C are 45.8 torr and 10.9 torr , respectively. Assuming ideal behavior, calculate each of the following. (Note that the mole fraction of an individual gas component in an ideal gas mixture can be expressed in terms of the component's partial pressure.)
The vapor pressure of heptane in the mixture.
The vapor pressure of octane in the mixture.
The total pressure above the solution.
What is the mass percent concentration of heptane in vapor phase?
What is the mass percent concentration of octane in vapor phase?
Explanation / Answer
The vapor pressure of heptane in the mixture.
The vapor pressure of octane in the mixture.
The total pressure above the solution.
What is the mass percent concentration of heptane in vapor phase?
What is the mass percent concentration of octane in vapor phase?
(43.0 g C7H16) / (100.2019 g C7H16/mol) = 0.42913 mol C7H16
(42.0 g C8H18) / (114.2285 g C8H18/mol) = 0.36768 mol C8H18
(45.8 torr C7H16) x (0.42913 mol C7H16) / (0.42913 mol + 0.36768 mol) = 24.666 torr C7H16
(10.9 torr C8H18) x (0.36768 mol C8H18) / (0.42913 mol + 0.36768 mol) = 5.0297 torr C8H18
=(5.0297 torr C8H18) / (5.0297 torr + 24.666 torr) = 0.16937 (mole fraction of C8H18 in the vapor)
1 - 0.16937 = 0.83063 (mole fraction of C7H16 in the vapor)
Take a hypothetical sample of exactly 1 mole of the vapor:
=(0.16937 mol C8H18) x (114.2285 g C8H18/mol) = 19.347 g C8H18
=(0.83063 mol C7H16) x (100.2019 g C7H16/mol) = 83.231 g C7H16
=(83.231 g C7H16) / (83.231 g + 19.347 g) = 0.81139 = 81.1% heptane by mass in the vapor
=100% - 81.1% = 18.9% octane by mass in the vapor
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