A solution contains 48.0 g of heptane ( C 7 H 16 ) and 57.0 g of octane ( C 8 H

Question

A solution contains 48.0g of heptane (C7H16) and 57.0g 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.)

A. The vapor pressure of heptane in the mixture.

B. The vapor pressure of octane in the mixture.

C. The total pressure above the solution.

D. What is the mass percent concentration of heptane in vapour phase?

E. What is the mass percent concentration of octane in vapour phase?

F. Why is the composition of the vapor different from the composition of the solution?

Explanation / Answer

. A solution contains 50.0 g of heptane (C7H16) and 50.0 g of octane (C8H18) at 250C. The vapor pressures of pure hexane and pure octane at 250C are 45.8 torr and 10.9 torr, respectively.

Assuming ideal behavior , answer the following:

a. What is the vapor pressure of each of the solution components in the mixture?

b. What is the total pressure above the solution?

c. What is the composition of the vapor in mass percent?

d. Why is the composition of the vapor different from the composition of the solution?

n heptane = 50 / 100 = 0.5 n octane = 50 / 114 = 0.439

a) X heptane = 0.5 / 0.939 = 0.532 X octane = 0.468

So P heptane = 0.532 * 45.8 = 24.3656 torr P octane = 0.468 * 10.9 = 5.1012 torr

b) total pressure = 23.3656 + 5.1012 = 28.4668 torr

c) X heptane in vapour = 23.3656 / 28.4668 = 0.821 X octane in vapour = 0.179

d) composition of vapour is different from solution because pure vapour pressure of hexane and octane are different .

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