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A natural product (MW=150) distills with steam at a boiling temperature of 99 degrees Celsius at atmospheric pressure. The vapor pressure of water at 99 degrees Celsius is 733 mm Hg.

A) Calculate the weight of the natural product that codistills with each gram of water at 99 degrees Celsius.

B) How much water must be removed by steam distillation to recover this natural product from 0.5 g of a spice that contains 10% of the desired substance?

I'll get you started.
P(H2O)/P(org) = n(H2O)/n(org). Let's simplify some by calling H2O=w and organic = o
Pw/Po = nw/no
The total pressure is 1 atm or 760 mm Hg.
Since the partial pressure of water at this temperature from the problem is 733 mm, then the partial pressure of the organic material must be 760-733 = 27 mm.
We also know that n = grams/molar mass.
Substituting we get
733/27 = (grams w/18)/(grams o/150).
The problem wants to know grams o for 1 g water so substitute 1 for grams w.
733/27 = (1/18)/grams o/150).
Solve for grams o and you will have the answer to part (A).

I solved for o and got 0.306957708 grams. Correct?

For B, I did this: 0.5 X 0.10 = 0.05 g

0.306957708 - 0.05 = 0.256957708 g of water, which is my answer. Correct?

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