Question

Pentane (C5H12) and hexane (C6H14) form an ideal solution. At 25.0 oC the vapor pressure of pure pentane and pure hexane are 511 and 150. Torr, respectively. A solution is prepared by mixing 30.0 mL pentane (density, 0.63 g/mL) with 50.0 mL hexane (density, 0.66 g/mL).

Required:
a. What is the vapor pressure of this solution?
b. What is the mole fraction of pentane in the vapor that is in equilibrium with this solution?

Answer 1

a. 297 Torr

b. 0.700 is mole fraction of pentane in the vapor

Step-by-step explanation:

a. The vapour pressure of a solution, P, is:

P = Pₐ°Xₐ+Pₙ°Xₙ

Where P° is vapour pressure of pure solvent and X its mole fraction.

To solve this question we need to find the moles of each liquid in order to find mole fraction and vapour pressure.

Moles pentane -Molar mass: 72.15g/mol-

30.0mL * (0.63g/mL) = 18.9g * (1mol/72.15g) = 0.262 moles pentane

Moles hexane-Molar mass: 72.15g/mol-

50.0mL * (0.66g/mL) = 33g* (1mol/86.18g) = 0.383 moles hexane

Mole fraction pentane:

0.262 moles pentane / 0.262 moles pentane+0.383 moles hexane

= 0.4062

Mole fraction hexane:

1 - 0.4062 = 0.5938

Vapour pressure:

P = Pₐ°Xₐ+Pₙ°Xₙ

P = 511Torr*0.4062+150Torr*0.5938

P = 297 Torr

b. The mole fraction times vapour pressure is the vapour pressure of the liquid. As pressure is directly proportional to moles, the mole fraction of pentane in the vapor phase is:

X = Pₐ°Xₐ / Pₐ°Xₐ+Pₙ°Xₙ

Where a is pentane and n hexane

X = 511Torr*0.4062 / 511Torr*0.4062+150Torr*0.5938

X = 0.700 is mole fraction of pentane in the vapor