The Calculation of Vapor Pressure in a Solution of Pentane and Hexane

Understanding the Vapor Pressure of Pentane and Hexane Solution

Pentane (C₅H₁₂) and hexane (C₆H₁₄) form an ideal solution. At 25ᵒC, the vapor pressures of pentane and hexane are 511 and 150 torr, respectively. A solution is prepared by mixing 25 mL of pentane (density, 0.63 g/mL) with 45 mL of hexane (density, 0.66 g/mL).

a. What is the vapor pressure of the resulting solution?

Final answer: The vapor pressure of the solution resulting from combining pentane and hexane is determined using Raoult's Law. The respective mole fractions of pentane and hexane in the solution are obtained first. Applying these values into Raoult's Law gives a total vapor pressure of approximately 292 torr.

Explanation: The vapor pressure of the resulting solution can be calculated using Raoult's law, which states that the partial pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture. First, we calculate the moles of pentane and hexane using the formula: moles = density * volume / molar mass.

For Pentane: molesPentane = (0.63g/mL * 25mL)/(72.15g/mol) = 0.218 mol

For Hexane: molesHexane = (0.66g/mL * 45mL)/(86.18g/mol) = 0.345 mol

Next, we calculate the mole fractions: mole fraction = moles of component / total moles. moleFractionPentane = 0.218 mol / (0.218 mol + 0.345 mol) = 0.387. MoleFractionHexane = 0.345 / (0.218 mol + 0.345 mol) = 0.613.

Lastly, we plug the mole fractions and the pure component vapor pressures into Raoult's law: Ptotal = Ppentane * moleFractionPentane + Phexane * moleFractionHexane.

Ptotal = (511 torr * 0.387) + (150 torr * 0.613)

Finally, the vapor pressure of the resulting solution is approximately 292 torr.