A solution contains 57.0 g of heptane (C7H16) and 50.0 g of octane (C8H18) at 25

Question

A solution contains 57.0 g of heptane (C7H16) and 50.0 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.)

1)Whats The vapor pressure of heptane in the mixture

2) The vapor pressure of octane in the mixture.

3)The total pressure above the solution.

4)What is the mass percent concentration of heptane in vapor phase?

5)What is the mass percent concentration of octane in vapor phase?

6)Why is the composition of the vapor different from the composition of the solution?

Explanation / Answer

Moles = mass/ molar mass

Atomic weights : C=12 and H=1

Molar masses: C7H16= 7*12+16=100, C8H18=8*12+18=114

Moles : Heptane= 57/100=0.57 and octane = 50/114=0.44

Total moles= 0.57+0.44=1.01

Mole fraction= moles of the component/total moles

Mole fractions : Heptane=0.57/1.01 =0.565 and octane- 0.44/1.01=0.435

Vapor pressure of component in the solution=moles fraction of that component* pure component vapor pressure

Vapor pressure in solution ( Torr) : heptane = 0.565*45.8 =26 Torr and octane =0.435*10.9 =4.7 Torr

Total pressure- Vapor pressure of heptane in solution + vapor pressure of octane in solution =26+4.7= 30.7 Torr

Mole fraction of the component in the vapor phase = vapor pressure of component in solution/total pressure

Mole fractions in the vapor phase : Heptane= 26/30.7=0.85 and octane= 1-0.85=0.15

When the liquid of given component is heated, the more volatile component gets enriched in the vapor phase and the less volatile component gets enriched in the liquid phase.

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