Question

A solution contains two isomers, n-propyl alcohol and isopropyl alcohol, at 25°C. The total vapor pressure is 38.6 torr. What are the mole fractions of each alcohol in the liquid and in the vapor phase? The vapor pressures are 21.0 torr for n-propyl alcohol and 45.2 torr for isopropyl alcohol.

Answer 1

Answer:

Mole fraction of alcohols in liquid phase $x_1=0.2727\& x_2=0.7273$.

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$.

Explanation:

The total vapor pressure of the solution = p =38.6 Torr

Partial vapor pressure of the  n-propyl alcohol =$p^{o}_1=21.0 Torr$

Partial vapor pressure of the isopropyl alcohol =$p^{o}_2=45.2 Torr$

$p=x_1\times p^{o}_1+x_2\times p^{o}_2$  (Raoult's Law)

$p=x_1\times p^{o}_1+(1-x_1)\times p^{o}_2$

$38.6 Torr=x_1\times 21.0 Torr+(1-x_1)\times 45.2 Torr$

$x_1=0.2727$

$x_2=1-0.2727=0.7273$

$x_1\& x_2$ is mole fraction in liquid phase.

Mole fraction of components in vapor phase $y_1\& y_2$

$p_1=y_1\times p$ (Dalton's law of partial pressure)

$y_1=\frac{p_1}{38.6 Torr}=\frac{p^{o}_1\times x_1}{38.6 Torr}$

$y_1=\frac{21.0 Torr\times 0.2727}{38.6 Torr}=0.1468$

$y_1=\frac{p_2}{38.6 Torr}=\frac{p^{o}_2\times x_2}{38.6 Torr}$

$y_2=\frac{45.2 Torr\times 0.7273}{38.6 Torr}=0.8516$

Mole fraction of alcohols in vapor phase $y_1=0.1468\& y_2=0.8516$