posted by Tim on .
Calculate the vapor pressure of water over each of the following ethylene glycol C2H6O2 solutions at 80 degree C (vp=pure water=657.6 mm Hg). Ethylene glycol can be assumed to be nonvolatile.
(a) X (ethylene glycol)= 0.288
(b) % ethylene glycol by mass =39%
(c) 2.42 m ethylene glycol
calculate a, b, c.
This is an exercise in Raoult's Law.
If you have not heard of it, review:
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You will need the Mole Fraction X of each mixture. In (2), they have told you that X = 0.288, so the answer is
0.288 * 657.6 mm Hg = _____
My answer was wrong for (a)
The mole fraction of water is Xh2o = 1 - 0.288 = 0.712. Use that value in the formula
Could you provide a little bit more clues i am still confused.
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