At 25°C the vapor pressures of pentane and hexane are 513 and 152 torr, respectively. Solution is prepared by mixing 23 mL of pentane (density = 0.63 g/mL) with 69 mL of hexane (density = 0.66 g/mL).

Question

Pentane (C5H12) and hexane (C6H14) combine to form an ideal solution. At 25°C the vapor pressures of pentane and hexane are 513 and 152 torr, respectively. A solution is prepared by mixing 23 mL of pentane (density = 0.63 g/mL) with 69 mL of hexane (density = 0.66 g/mL).

(a) What is the vapor pressure (in torr) of this solution?

(b) What is the mole fraction of pentane in the vapor that is in equilibrium with this solution?

Answer 1

This is a mixture of two VOLATILE solutes, and so each will affect the vapor pressure in relation to the mole fraction present. So, the first thing to do is to calculate the mole fraction (X) of each.

Pentane: 23 ml x 1 L/1000 ml x 0.63 g/ml x 1 mol/72.15 g = 2x10^-4 moles pentane

Hexane: 69 ml x 1 L/1000 ml x 0.66 g/ml x 1 mol/86.2 g = 5.3x10^-4 moles hexane

Total moles = 7.3x10^-4 moles

X_pentane = 2/7.3 = 0.274

X_hexane = 5.3/7.3 = 0.726

a) Vapor pressure of solution = (0.274 x 513 torr) + (0.726 x 152 torr) = 141 + 110 = 251 torr

b) X_pentane in vapor = 110/251 = 0.438