Limone = L
water = W
pL = partial pressure limone
pW = partial pressure W
ML = molar mass Limone
MW = molar mass Water
---------------------
(mass L/mass W) = (pL*ML/pW x MW)
pTotal = pL + pW and pW = 690 from the problem and pTotal = 760.
Solve for the mass ratio and convert to percent L. Post your work (I suggest a new post at the top but include this information--just repost this if you need further assistance.)
Limone (MW 136g/mol) is a pleasant smelling liquid found in lemon and orange peels. The boiling point of limonene is 175 ˚C, but it co-distills with water at 97.5 ˚C. If the vapour pressure of water at 97.5˚C is 690 mmHg, what percent (by mass) of the steam-distillate is limonene? (Assuming that the distillation is occurring at 760 mmHg)
2 answers
89