Question

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

Answer 1

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 .