Question

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?

Answer 1

The mole fraction of cyclohexane in the liquid phase is 36.8%.

Step-by-step explanation:

To find the mole fraction of cyclohexane in the liquid phase, we need to calculate the total moles of cyclohexane and 2-methylpentane in the mixture. First, we calculate the moles of each component using their molar masses:

moles of cyclohexane = 25 g / 84.18 g/mol = 0.297 mol

moles of 2-methylpentane = 44 g / 86.18 g/mol = 0.509 mol

Next, we calculate the mole fraction of cyclohexane:

mole fraction of cyclohexane = moles of cyclohexane / (moles of cyclohexane + moles of 2-methylpentane)

mole fraction of cyclohexane = 0.297 mol / (0.297 mol + 0.509 mol) = 0.368 or 36.8%

Answer 2

The mol fraction of cyclohexane in the liquid phase is 0.368

Step-by-step 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