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, Xcy(vapor), is cyclohexane?

Answer 1

This is an incomplete question. The complete question is as : Part B :A solution is composed of 1.60 mol cyclohexane (97.6 torr) and 2.80 mol acetone (229.5 torr). What is the total vapor pressure above this solution?

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,Xcy(vapor) , is cyclohexane?

Answer:  The mole fraction of cyclohexane in the vapor above the solution is 0.195

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_1=x_1p_1^0$ and $p_2=x_2p_2^0$

where, x = mole fraction in solution  

$p^0$ = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2$

$p_{total}=x_Ap_A^0+x_Bp_B^0$

$x_{cyclohexane}=\frac{\text {moles of cyclohexane}}{\text {moles of cyclohexane+moles of acetone}}=\frac{1.60}{1.60+2.80}=0.364$,  

$x_{acetone}=1-x_{cyclohexane}=1-0.364=0.636$,  

$p_{cyclohexane}^0=97.6torr$

$p_{acetone}^0=229.5torr$

$p_{total}=0.364\times 97.6+0.636\times 229.5=181torr$

$y_{cyclohexane}$ = mole fraction of cyclohexane in vapor phase $y_{cyclohexane}=\frac{p_{cyclohexane}}{p_{total}}=\frac{0.363\times 97.6}{182}={35.4}{181}=0.195$

Thus the mole fraction of cyclohexane in the vapor above the solution is 0.195