A solution is made by adding 0.100 mole of ethyl ether to 0.841 mole of ethyl alcohol. If the vapor pressure of ethyl ether and ethyl alcohol at 20oC are 375 torr and 20.0 torr, respectively, the vapor pressure of the solution at 20oC (assuming ideal behavior) is:
Before I forget it, did you get the problem of toluene and benzene, in which the problem asked for X of one of them IN THE SOLUTION, worked? I can do that one if you post it.
moles ether = ?
moles EtOH = ?
Xether = nether/total mols.
XEtOH = nEtOH/total mols.
pether = Xether*Poether
pEtOH = XEtOH*PoEtOH
Then total vapor pressure = pethr + pEtOH.
To find the vapor pressure of the solution, we need to use Raoult's Law, which states that the partial pressure of a component in a solution is proportional to the mole fraction of that component.
The mole fraction is calculated by dividing the moles of a component by the total moles of all components in the solution.
Given:
Moles of ethyl ether (C2H5OC2H5) = 0.100 mol
Moles of ethyl alcohol (C2H5OH) = 0.841 mol
The total moles of the solution is the sum of the moles of ethyl ether and ethyl alcohol.
Total moles = 0.100 mol + 0.841 mol = 0.941 mol
Now, we can calculate the mole fraction of each component.
Mole fraction of ethyl ether (Xether) = Moles of ethyl ether / Total moles
= 0.100 mol / 0.941 mol
Mole fraction of ethyl alcohol (Xalcohol) = Moles of ethyl alcohol / Total moles
= 0.841 mol / 0.941 mol
Next, we need to find the vapor pressure of each component in the solution using Raoult's Law:
Partial pressure of ethyl ether (Pether) = Mole fraction of ethyl ether * Vapor pressure of ethyl ether
= Xether * 375 torr
Partial pressure of ethyl alcohol (Palcohol) = Mole fraction of ethyl alcohol * Vapor pressure of ethyl alcohol
= Xalcohol * 20.0 torr
Now, we can substitute the values and calculate the vapor pressure of the solution:
Vapor pressure of the solution = Pether + Palcohol
Finally, solve the equation to get the answer.