At 63.5 degreesC, the vapour pressure of H2O is 23.3kPa and that of ethanol is 53.3kPa. A solution is made by mixing egual masses of H2O and C2H5OH.
a. what is the mole fraction of ethanol in the solution?
b. assuming ideal-solution behaviour, what is the vapour pressure of the solution at 63.5 degreeC.
c. what is the mole fraction of ethanol in the vapour above the solution.
Read this:
http://en.wikipedia.org/wiki/Raoult%27s_law
thenkyou. i missed reading that pert on that page before. i knew it was something simple.
i now have parts A and B but how do you work out C? i have the mole fraction of ethanol from part A. why is it not the same?
The mole fraction of the LIQUID is moles liquid component/total moles liquid.
The mole fraction of the VAPOR is moles vapor component/total vapor moles. Since the vapor pressures are not the same the vapor is more concd in the more volatile component. That's one reason fractional distillation works.
so do i use the moles i would have from part A? surely not, becayse they just give me the mole fraction of the solution.
what do i use? i have the answer of 0.472- don't know how to get it.
i feel so dumb and have my exam on 23rd.
To find the answers to the given questions, we need to apply the concepts of vapour pressure and mole fraction. Let's solve each part step by step.
a. To find the mole fraction of ethanol in the solution, we need to first calculate the moles of each component (H2O and C2H5OH) and then divide the moles of ethanol by the total number of moles.
Given:
Vapour pressure of H2O (PH2O) = 23.3 kPa
Vapour pressure of ethanol (PC2H5OH) = 53.3 kPa
Since equal masses of H2O and C2H5OH are mixed, we can assume that the mole ratio is also 1:1. Therefore, we need to compare the vapour pressure ratios and use Raoult's law, which states that the vapour pressure of a solution is equal to the mole fraction of each component multiplied by its pure vapour pressure.
Mole fraction of ethanol (XEtOH) = Moles of ethanol / Total moles
First, let's find the moles of each component:
Moles of H2O = mass of H2O / molar mass of H2O
Moles of C2H5OH = mass of C2H5OH / molar mass of C2H5OH
Since the masses of H2O and C2H5OH are equal, we can use the same mass value for both calculations.
Now, substitute the mole values into the equation to find the mole fraction of ethanol.
b. To calculate the vapour pressure of the solution at 63.5 degrees Celsius, we can use Raoult's law, which states that the total vapour pressure of a solution is equal to the sum of the vapour pressures of each component multiplied by their respective mole fractions.
Vapour pressure of the solution (PSolution) = (XH2O * PH2O) + (XEtOH * PC2H5OH)
Substitute the mole fraction of each component and their corresponding vapour pressures into the equation to find the pressure of the solution.
c. To find the mole fraction of ethanol in the vapour above the solution, we can use Dalton's law of partial pressures. According to this law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of each component in the mixture.
To calculate the mole fraction of ethanol in the vapour, we need to consider the partial pressure of ethanol (PEtOH) and the total pressure of the mixture (PSolution), which we calculated in part b.
Mole fraction of ethanol in the vapour (XEtOHvapour) = PEtOH / PSolution
Substitute the partial pressure of ethanol and the total pressure of the mixture into the equation to find the mole fraction of ethanol in the vapour.
By following these steps, you should be able to calculate the mole fraction of ethanol in the solution, the vapour pressure of the solution, and the mole fraction of ethanol in the vapour above the solution.