# 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?

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Some data necessary to solve this:

Methanol #MM = (32.0g)/(mol)#

Vapor pressure #=717.2t o rr#

Ethanol: #MM = (46.1g)/(mol)#

Vapor pressure #=400.0t o rr#

Thank you!

Some data necessary to solve this:

Methanol

Vapor pressure

Ethanol:

Vapor pressure

Thank you!

##### 1 Answer

#### Answer:

#### 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

#"0.78125 moles + 1.6269 moles = 2.40815 moles"#

Now, according to Raoult's Law, the **partial pressure** of methanol above this solution,

#P_"MetOH" = chi_ "MetOH" * P_"MetOH"^@#

Here

#chi_"MetOH"# is themole fractionof methanol in the mixture#P_"MetOH"^@# is the vapor pressure ofpure methanolat#63.5^@"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_"MetOH" = 0.3244 * "717.2 torr"#

#P_"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_"EtOH" = 0.6756 * "400.0 torr"#

#P_"EtOH" = "270.24 torr"#

Now, the **total pressure** above the solution is equal to

#P_"total" = P_"MetOH" + P_"EtOH"#

#P_"total" = "232.66 torr + 270.24 torr"#

#P_"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_"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**.