What is the mole fraction of methanol CH3OH, in the vapor phase from a solution consisting of a mixture of methanol and propanol CH3CH2CH2OH at 40celsius, if the total pressure above the solution is 131 torr? The vapor pressures of methanol and propanolare 303 and 44.60 torr respectively. Assuming these compounds form nearly ideal solutions when mixed together.
Propanol = pr
Methanol = me
Pure pr = 44.6
Pure me = 303
The difference between pure pr and pure me is 303-44.6 = 258.4
delta P from pure pr = 131-44.6 = 86.4
delta P from pure me = 303-131 = 172.0
86.4/258.4 = mole fraction me.
172.0/258.4 = mole fraction pr.
The best way to see this is to draw it out on a piece of paper.
|........131....................|
44.6.....x.....................303
To check the answer I did this.
0.3344 x 303 = 101.32 mm
0.6656 x 44.6 = 29.68 mm
Total 131.000. Of course, that many significant figures are not allowed. You would round to 0.334 and 0.666 which gives 130.9 which rounds to 131. I hope this helps.
I followed this procedure but i was told the answer was incorrect. Any help would be appreciated.
The above procedure is correct as far as I went BUT the numbers calculate the mole fraction of me and pr in the liquid phase. To find the mole fraction of each in the vapor phase,
Xme vapor phase = Pme/total P
Xpr vapor phase = Ppr/total P.
To find the mole fraction of methanol (CH3OH) in the vapor phase from the given solution, we start by understanding the concept of mole fraction.
Mole fraction (Xi) is defined as the ratio of the number of moles of a component to the total number of moles of all components in a mixture. It is usually denoted by the symbol Xi.
The total pressure above the solution is given as 131 torr. The vapor pressures of methanol and propanol are 303 and 44.60 torr, respectively.
First, we need to calculate the partial pressure of methanol (Pm) in the vapor phase.
According to Raoult's law, the partial pressure of a component in an ideal solution can be calculated by multiplying the mole fraction of that component by its vapor pressure at that temperature.
Let's assume Xmethanol represents the mole fraction of methanol in the solution.
Pm = Xmethanol * Pmethanol
Since the pressure is given in torr, it's important to convert the pressure to atm since mole fraction is based on the pressure in atm.
1 atm = 760 torr
Pm = Xmethanol * (Pmethanol / 760) * 1 atm
Given:
Pmethanol = 303 torr
Pm = 131 torr
We can rearrange the equation to solve for Xmethanol:
Xmethanol = (Pm * 760) / Pmethanol
Now, let's substitute the values and calculate Xmethanol:
Xmethanol = (131 * 760) / 303
Xmethanol ≈ 328.976
However, mole fraction is a dimensionless quantity and should range between 0 and 1. Therefore, we need to normalize the value by dividing it by the sum of mole fractions of all components.
To do that, we need to calculate the mole fraction of propanol (CH3CH2CH2OH) as well.
Xpropanol = 1 - Xmethanol
Now, we can normalize the mole fractions:
Xmethanol_normalized = Xmethanol / (Xmethanol + Xpropanol)
Xpropanol_normalized = Xpropanol / (Xmethanol + Xpropanol)
Substituting the calculated values:
Xpropanol_normalized = (1 - 0.328976) / (0.328976 + 1- 0.328976)
= 0.671024 / 1.328976
≈ 0.505
Xmethanol_normalized = 1 - Xpropanol_normalized
= 1 - 0.505
≈ 0.495
Therefore, the normalized mole fraction of methanol in the vapor phase from the given solution is approximately 0.495.