A naturally occurring oil co-distills with water to produce an oil/water distillate that is 20% oil by weight. If the molecular
weight of the oil 100 g/mol, what was the partial pressure of the oil during distillation assuming atmospheric pressure is 760 mm Hg
1 answer:
Answer:
Explanation:
Partial pressure of oil = mole fraction of oil x total pressure
mole fraction of oil = mole of oil / mole of water + mole of oil
= mole of oil = mass of oil / molecular weight of oil
= 20 / 100 = .2
mole of water = 80 / 18
= 4.444
mole fraction of oil = .2 / .2 + 4.444
= .2 / 4.644
Partial pressure of oil = mole fraction of oil x total pressure
= (.2 / 4.644 ) x 760 mm
= 32.73 mm Hg .
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