At 99˚, the vapor pressure of water is 733 mm Hg.
(a) At standard atmospheric pressure, what is the vapor pressure of a compound being steam-distilled at this temperature?
(b) If the compound has a molecular weight of 180, how much water is required to distill 1.0 g of the compound?

Question

Quizplus · Accepted Answer

The Answer is (a) P_total= P˚ _A+ P˚ _B
If P˚ _A(water) is 733 mm and P_totalis 760, then P˚ _B(the compound being distilled) is (760 - 733) or 27 mm.
(b) (wt_A/ wt_B) = (P˚ _A/ P˚ _B) x (MW_A/ MW_B)
wt_A= 1.0 x (733 / 27) x (18 / 180) = 2.7 g of water

At 99˚, the Vapor Pressure of Water Is 733 Mm