Question

At 300 K, the vapor pressure of pure benzene (C6H6) is 0.1355 atm and the vapor pressure of pure n-hexane (C6H14) is 0.2128 atm. Mixing 50.0 g of benzene with 50.0 g of n-hexane gives a solution that is nearly ideal. (a) Calculate the mole fraction of benzene in the solution. (b) Calculate the total vapor pressure of the solution at 300 K. (c) Calculate the mole fraction of benzene in the vapor in equilibrium with the solution.

Answer 1

Answer:

A. 0525.

B. 0.3483 atm.

C. 0.3890.

Explanation:

A.

Molar mass of Benzene =

(12*6) + (1*6)

= 78 g/mol.

Number of moles of Benzene = mass/molar mass

= 50/78

= 0.641 moles.

Molar mass of n-hexane =

(12*6) + (1*14)

= 86 g/mol.

Number of moles of n-hexane = mass/molar mass

= 50/86

= 0.581 moles.

Total moles of the solution = number of moles of Benzene + number of moles of n-hexane

= 0.641 + 0.581

= 1.222 moles

Mole fraction is the number of moles of a particular substance in a solution divided by the total number of moles of substances in the solution.

Mole fraction of Benzene = 0.641/1.222

= 0.525.

B.

Vapour Pressure of a solution is the amount of pressure that the vapour exert on the liquid solvent when they are in equilibrium and at a certain temperature.

Uaing Raouit's law and Dalton's law,

Raoult's law states that the the partial pressure of a solution is directly proportional to the mole fraction of the solute component.

Dalton's law states that the total pressure of a solution is the sum of its individual partial pressure.

Uaing Dalton's law,

Psol = Pbenzene + Pn-hexane

= 0.1355 + 0.2128

= 0.3483 atm

C.

Using Raoult's equation, Mole fraction of Benzene = P°benzene/Psol

= 0.1355/0.3483

= 0.3890