vapor pressure calculation

[Smooth Operater]

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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 is added to 200 gram of water, what is the approxmate vapor pressure of the mixture?

the answer is C, but I am not sure how would you arrive to that? Thanks!

 

[jhuplaw]

wouldn't you just find the partial pressures of water and ethanol then add them together?

what was answer choice C by the way?

 

[doc toothache]

Mole fraction of water is 200/250=0.8
Mole fraction of ethanol is 50/250=0.2

total pressure 0.2(58.9) + 0.8(23.8)= 30.82

 

[Hyphae]

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 is added to 200 gram of water, what is the approxmate vapor pressure of the mixture?

the answer is C, but I am not sure how would you arrive to that? Thanks!

The VP that is given, is for PURE ethanol and PURE water

VP(ethanol) = VP(pure ethanol) x Mole Fraction of Solvent (in this case is water)

Mole Fraction = moles of (water) / total moles (ethanol + water)

VP (mixture) = VP(ethanol) + VP(water) = (58.9)(mole fraction[solvent]) + (23.8)(mole fraction[solvent])

And I don't think toothache's mole fraction is correct but it might be an approx.

 

[doc toothache]

Hyphae is right, the figures are incorrect. They only represent percentage composition. You are on the right track, however.....

Moles of H20= g water/mol. wgt=200/18= 11.11
Moles of EtOH=g EtOH/mol. wgt=50/46= 1.09

Total moles of H2O and EtOH =11.11=1.09=12.2

Mole fraction H2O= moles H20/moles H2O+moles EtOH = 11.11/12.2=0.91
Mole fraction EtOH=moles EtOH/moles H20+moles EtOH= 1.09/12.2=0.09

VP(mixture)=VP H2O x mole fraction H2O + VP EtOH x mole fraction EtOH

VP(mixture)=23.8(0.91)+58.9(0.09)
VP(mixture)=21.66+5.20
VP(mixture)=26.96