A solution contains 49.0 g of heptane (C7H16)and 59.0 g of octane (C8H18) at 25

Question

A solution contains 49.0 g of heptane (C7H16)and 59.0 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.)

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 percent concentration of heptane in vapor phase

E) What is the percent concentration of octane in vapor phase

F) Why is the composition of the vapor different from the composition of the solution

Explanation / Answer

Molar mass of Heptane (C7H16) = 7 * 12 + 16 * 1 = 84 + 16 = 100 gm/mol

Number of moles of Heptane = 49/100 = 0.49

Molar mass of Octane(C8H18) = 8 * 12 + 18 * 1 = 114 gm/mol

NUmber of moles of Octane = 59/114 = 0.5175 mole

Mole fraction of Heptane = 0.49/(0.49+0.5175) = 0.48635

Mole fraction of Octane = 1 - 0.48635 = 0.51364

a) Vapor pressure of Heptane in mixture = mole fraction of Heptane * vapor pressure of pure heptane

=> 0.48635 * 45.8 = 22.274 torr

b) Vapor pressure of Octane in mixture = mole fraction of Octane * vapor pressure of pure Octane

=> 0.51364 * 10.9 = 5.598 torr

c) Total Vapor Pressure = 22.274 + 5.598 = 27.872 torr

d) Percent concentration of Heptane = 48.36%

e) Percent concentration of Octane = 51.36

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