As you saw in Part B, the vapor above the cyclohexane-acetone solution is compos

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_{ m cy}( m vapor)), is cyclohexane?

Part B:

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

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.