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Topic: Steam Diistillation and Vapor Pressure  (Read 4764 times)

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Offline Sis290025

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Steam Diistillation and Vapor Pressure
« on: December 04, 2006, 10:17:50 AM »
A natural product (MW = 150 g/mol) distills with steam at a boiling temperature of 99 C at atmospheric pressure. The vapor pressure of water at 99 C is 733 mmHg.

a. Calculate the weight of the natural product that codistills with each gram of water at 99 C.


(P*_a)/(P*_b) = moles A / moles B
P*_a = 760 torr
P*_b = 733 torr

ratio  760 moles/733 moles water =  1.0368 mol/1 mol water
 1.0368 mol (150 g/mol) = 155.52 g organic compound / (18.0 g water) = 8.64 g/ 1 g water ?

 

b. How much H2O must be removed by steam distillation to recover this natural product from 0.5 g of spice that contains 10 % of the desired substance?

5 g*(.10) = .05 g desired substance

0.05 g (1 g water/8.64 g organic) = 0.00579 g water ???

Thanks. 

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