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?
Correct Answer
verified
(a) Ptotal = P˚ A + P˚ B
If P˚ A (water) is 733 mm and Ptotal is 760, then P˚ B (the compound being distilled) is (760 - 733) or 27 mm.
(b) (wtA/ wtB) = (P˚ A/ P˚ B) x (MWA/ MWB)
wtA = 1.0 x (733 / 27) x (18 / 180) = 2.7 g of water
At 99.6˚, water has a vapor pressure of 750 mm Hg and quinoline (immiscible with water) has a vapor pressure of 10 mm Hg. What weight of water must be distilled for each gram of quin- oline in a steam distillation at 760 mm Hg?
Aniline (MW 93) co-distills with water at 98.2˚. The partial vapor pressure of water at 98.2˚ is 712 mm Hg. You need to isolate 10 g of aniline. What is the minimum amount of water it will take to steam-distill this amount of aniline?
A mixture of immiscible liquids (both water-insoluble) is subjected to steam distillation. At 90˚, the vapor pressure of pure water is 526 mm Hg. If the vapor pressure of compound A is 127 mm Hg and that of B is 246 mm Hg at 90˚:
(a) what is the total vapor pressure of the mixture at 90˚?
(b) would this mixture boil at a temperature above or below 90˚?
(c) what would be the effect on the vapor pressure and boiling temperature by doubling the amount of water used?
Limonene (MW 136) is a pleasant-smelling liquid found in lemon and orange peels. The bp of limonene is 175˚, but it co-distills with water at 97.5˚. If the vapor pressure of water at 97.5˚ is 690 mm Hg, what percentage of the steam-distillate is limonene?