GC vapor pressure question

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At 25 celsius, water has a vapor pressure of 23.8 torr and ethanol has a vapor pressure of 58.9 torr. IF 50 grams of ethanol (MW = 46) is added to 200 grams of water, what is the approximate vapor pressure of the mixture?
 
The vapor above the mixture will consist of both ethanol and water molecules because both components of the mixture are volatile in this case. Therefore, Vapor pressure of the mixture=vapor pressure of ethanol above the mixture + vapor pressure of water above the mixture. Now, simply use the vapor pressure lowering equation (X*Po=P) for ethanol, then water, and add them together. Po, the vapor pressure of pure water or pure ethanol, is given in the problem. X, the mole fraction, is solvable since you know the mass values of ethanol and water.
 
VP(H2O): 23.8 torr VPmixture= VP(H2O above solution)+ VP(Ethanol above solution)
VP(ethanol): 58.9 torr VPsolution= X(mole fraction) * VP (of pure water or ethanol)

mass (ethanol): 50g--> moles= 50/46= ~ 1 mol
mass (H2O): 200g--> moles= 200/18 = ~ 10 mol
X(ethanol)= 1/11(total moles)
X(H2O)= 10/11
VPsolution above H2O= (1/11)*(60 torr)= ~ 5.5
VPsolution above ethanol = (10/11)*(24 torr)= ~ 22

Now add the two together--> 5.5+22= Answer: ~27.5

Is that the right answer??? i did this really fast so there might be some calculation error
hope this helps =)
 
Is vapor pressure independent of volume? So 1 mL of water has the same vapor pressure as 500mL?
Thanks

VP(H2O): 23.8 torr VPmixture= VP(H2O above solution)+ VP(Ethanol above solution)
VP(ethanol): 58.9 torr VPsolution= X(mole fraction) * VP (of pure water or ethanol)

mass (ethanol): 50g--> moles= 50/46= ~ 1 mol
mass (H2O): 200g--> moles= 200/18 = ~ 10 mol
X(ethanol)= 1/11(total moles)
X(H2O)= 10/11
VPsolution above H2O= (1/11)*(60 torr)= ~ 5.5
VPsolution above ethanol = (10/11)*(24 torr)= ~ 22

Now add the two together--> 5.5+22= Answer: ~27.5

Is that the right answer??? i did this really fast so there might be some calculation error
hope this helps =)
 
ummm from what i know: vapor pressure is only dependent on the temperature. It is independent of surface area of the liquid, volume of the container, or the amount of liquid in the container...
so yes, 1 ml or 500ml of water, both will have the same vapor pressure...
 
You got it. I just figured it out by conceptual way from multiple choices, but I wanted to see how to get the same result by calculation. Thanks
 
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