A mixture contains 25 g of cyclohexane (C6H12) and 44 g of 2-methylpentane (C6H14). The mixture of liquids is at 35 oC. At this temperature, the vapor pressure of pure cyclohexane is 150 torr, and that of pure 2-methylpentane is 313 torr. Assume this is an ideal solution. What is the mole fraction of cyclohexane in the liquid phase?

The mol fraction of cyclohexane in the liquid phase is 0.368

Explanation:

Step 1: Data given

Mass of cyclohexane = 25.0 grams

Mass of 2-methylpentane = 44.0 grams

Temperature = 35.0 °C

The pressure of cyclohexane = 150 torr

The pressure of 2-methylpentane = 313 torr

The pressure we only need for the mole fraction in gas phase.

Step 2: Calculate moles of cyclohexane

Moles cyclohexane = mass cyclohexane / molar mass

Moles cyclohexane = 25.0 g / 84 g/mol = 0.298 mol of cyclohexane

Step 3: Calculate moles of 2-methylpentane

Moles = 44.0 grams / 86 g/mol = 0.512 mol of 2-methylpentane

Step 4: Calculate mole fraction of cyclohexane in the liquid phase

Mole fraction of C6H12:

0.298 / (0.298 + 0.512) = 0.368

The mol fraction of cyclohexane in the liquid phase is 0.368