# Calculate the mole fraction of methanol in the vapor phase at 63.5°C for an ideal solution containing 25.0g of pure methanol and 75.0g of pure ethanol?

## Some data necessary to solve this: Methanol $M M = \frac{32.0 g}{m o l}$ Vapor pressure $= 717.2 t o r r$ Ethanol: $M M = \frac{46.1 g}{m o l}$ Vapor pressure $= 400.0 t o r r$ Thank you!

Sep 16, 2017

$0.463$

#### Explanation:

Your tool of choice here will be Raoult's Law, which states that at a given temperature, the vapor pressure of a volatile component in an ideal mixture is equal to the mole fraction of the component in the mixture multiplied by the vapor pressure of the pure component.

This basically means that for an ideal mixture, the number of moles of each component of the mixture present in the solution will determine the vapor pressure of the component above the solution.

In your case, the solution contains

25.0 color(red)(cancel(color(black)("g"))) * "1 mole MetOH"/(32.0color(red)(cancel(color(black)("g")))) = "0.78125 moles MetOH"

75.0color(red)(cancel(color(black)("g"))) * "1 mole EtOH"/(46.1color(red)(cancel(color(black)("g")))) = "1.6269 moles EtOH"

The total number of moles present in the solution is equal to

$\text{0.78125 moles + 1.6269 moles = 2.40815 moles}$

Now, according to Raoult's Law, the partial pressure of methanol above this solution, ${P}_{\text{MetOH}}$, is given by

${P}_{\text{MetOH" = chi_ "MetOH" * P_"MetOH}}^{\circ}$

Here

• ${\chi}_{\text{MetOH}}$ is the mole fraction of methanol in the mixture
• ${P}_{\text{MetOH}}^{\circ}$ is the vapor pressure of pure methanol at ${63.5}^{\circ} \text{C}$

The mole fraction of methanol in the solution is equal to

chi_"MetOH" = (0.78125 color(red)(cancel(color(black)("moles"))))/(2.40815 color(red)(cancel(color(black)("moles")))) = 0.3244

The partial pressure of methanol above the solution is equal to

${P}_{\text{MetOH" = 0.3244 * "717.2 torr}}$

${P}_{\text{MetOH" = "232.66 torr}}$

Next, use the same approach to find the partial pressure of ethanol above the solution.

chi_"EtOH" = (1.6269 color(red)(cancel(color(black)("moles"))))/(2.40815color(red)(cancel(color(black)("moles")))) = 0.6756

You will have

${P}_{\text{EtOH" = 0.6756 * "400.0 torr}}$

${P}_{\text{EtOH" = "270.24 torr}}$

Now, the total pressure above the solution is equal to

${P}_{\text{total" = P_"MetOH" + P_"EtOH}}$

${P}_{\text{total" = "232.66 torr + 270.24 torr}}$

${P}_{\text{total" = "502.9 torr}}$

In order to find the mole fraction of methanol in the vapor phase, i.e. above the solution, use the fact that the partial pressure of methanol above the solution is equal to the mole fraction of methanol above the solution multiplied by the total pressure above the solution.

${P}_{\text{MetOH" = chi_"MetOH vapor" * P_"total}}$

This means that you have

chi_"MetOH vapor" = (232.66 color(red)(cancel(color(black)("torr"))))/(502.9color(red)(cancel(color(black)("torr")))) = color(darkgreen)(ul(color(black)(0.463)))

The answer is rounded to three sig figs.