Question

As you saw in Part B, the vapor above the cyclohexane-acetone solution is composed of both cyclohexane vapor and acetone vapor. What mole fraction of the vapor above the solution, \(X_{\rm cy}(\rm vapor)\), is cyclohexane?

Part B:

A solution is composed of 1.60 {\rm mol} cyclohexane (P_{\rm cy}^\circ=97.6~ \rm torr ) and 2.40 {\rm mol} acetone (P_{\rm ac}^\circ=229.5~ \rm torr ). What is the total vapor pressure P_{\rm total} above this solution?

Mole fraction of Cyclohexane = 1.60/(1.60+2.40) = 0.40

Mole fraction of Acetone = 2.40/(2.40+1.60) = 0.60

Using the raoults law

Total vapor pressure = mole fraction of cyclohexane * pressure of cyclohexane + mole fraction of acetone * pressure of acetone

=> 0.40 * 97.6 + 0.60 * 229.5

=> 176.74 torr

Answer 1

we know that

partial pressure of cyclohexane = mol fraction x pressure of pure cyclohexane

given

mol fraction of cyclohexane = 0.4

pressure of pure cyclohexane = 97.6

so

partial pressure of cyclohexane = 0.4 x 97.6 = 39.04

now

mol fraction of cyclohexane in vapor = partial pressure of cyclohexane / total pressure

given

total pressure = 176.74

so

mol fraction of cyclohexane in vapor = 39.04 / 176.74

mol fraction of cyclohexane in vapor = 0.221

so

mole fraction of cyclohexane in vapor is 0.221