Question

1-propanol (P1° = 20.9 Torr at 25 °C) and 2-propanol (P2° = 45.2 Torr at 25 °C) form ideal solutions in all proportions. Let x1 and x2 represent the mole fractions of 1-propanol and 2-propanol in a liquid mixture, respectively, and y1 and y2 represent the mole fractions of each in the vapor phase. For a solution of these liquids with x1 = 0.790, calculate the composition of the vapor phase at 25 °C.

Answer 1

Answer – We are given, mole fraction of 1-propanol, x1 = 0.790 , P₁° = 20.9 Torr at 25 °C , P₂° = 45.2 Torr at 25 °C

So, mole fraction of the 2-propanol, x2 = 1-0.790

                                                       = 0.210

We know,

Partial pressure = mole fraction * vapor pressure

For, partial pressure of 1-propanol = 0.790 *20.9 torr

                                                          = 16.5 torr

partial pressure of 2-propanol = 0.210 *45.2 torr

                                                = 9.50 torr

So the mole fraction of vapor 1-propanol = 16.5 torr / 16.5+9.50 torr

                                                                  = 0.635

mole fraction of vapor 2-propanol = 9.50 torr / 16.5+9.50 torr

                                                       = 0.365