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

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`.

Step-by-step 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* p^(o)_1+x_2* p^(o)_2` (Raoult's Law)

`p=x_1* p^(o)_1+(1-x_1)* p^(o)_2`

`38.6 Torr=x_1* 21.0 Torr+(1-x_1)* 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* p` (Dalton's law of partial pressure)

`y_1=(p_1)/(38.6 Torr)=(p^(o)_1* x_1)/(38.6 Torr)`

`y_1=(21.0 Torr* 0.2727)/(38.6 Torr)=0.1468`

`y_1=(p_2)/(38.6 Torr)=(p^(o)_2* x_2)/(38.6 Torr)`

`y_2=(45.2 Torr* 0.7273)/(38.6 Torr)=0.8516`

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