Question

Question #9bc92

Answer 1

Mole fraction of pentane: 0.45.

Explanation:

So, you know that you're dealing with a solution that contains pentane and hexane in a `1:4` mole ratio.

You also know that the vapor pressure of the two pure hydrocarbons at `20^@"C"` are 400 mmHg and 120 mmHg, respectively.

This means that you can use Raoult's Law to determine what the vapor pressure of the mixture will be. According to Raoult's Law, the partial pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

Moreover, the vapor pressure of the solution is equal to the sum of the individual vapor pressures, in this context called partial pressures, of its components.

So, the partial pressure of pentane and of hexane will be

`P_p = chi_p * P_p^@`

and

`P_h = chi_h * P_h^@`

The mole fraction of a substance that's part of a mixture is defined as the number of moles of that substance divided by the total number of moles present in the mixture.

This means that you have

`chi_p = n_p/(n_p + n_h)` and `chi_h = n_h/(n_p + n_h)`

Use the aforementioned mole ratio to find a relationship between `n_p` and `n_h`.

`n_p/n_h = 1/4 => n_h = 4 * n_p`

The two mole fractions become

`chi_n = n_p/(n_p + 4n_p) = cancel(n_p)/(5cancel(n_p)) = 1/5` and

`chi_h = (4n_p)/(n_p + 4n_p) = (4cancel(n_p))/(5cancel(n_p)) = 4/5`

The total vapor pressure of the mixture will thus be

`P_"total" = P_p + P_h`

`P_"total" = underbrace(1/5 * "400 mmHg")_(color(blue)(P_p)) + underbrace(4/5 * "120 mmHg")_(color(blue)(P_h)) = "176 mmHg"`

To get the mole fraction of pentane in the vapor phase, use the fact that you know the vapor pressure of pentane in solution.

`underbrace(P_"p solution")_(color(blue)(P_p)) = chi_"p vapor" * P_"total"`

`P_p = chi_"p vapor" * P_"total" => chi_"p vapor" = P_p/P_"total"`

`chi_"p vapor" = (80cancel("mmHg"))/(176cancel("mmHg")) = color(green)(0.45)`

SIDE NOTE I left the answer rounded to two sig figs.