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Topic: Vapor Pressure and Boiling Point - Check my Work  (Read 4010 times)

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Offline ozzlomo

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Vapor Pressure and Boiling Point - Check my Work
« on: May 04, 2009, 10:29:55 PM »
Calculate how much the vapor pressure lowers as a result of dissolving 20.2g of sucrose (C12H22O11) in 400.0 grams of water at 25 degrees C. What is the vapor pressure of the solution? The vapor pressure of pure water at this temperature is 23.76 torr. What is the vapor pressure at 100 degrees C?

Change in vapor pressure (atm):

 :delta: p = X2 Pdegrees

X 2 :

20.2g C 12 H 22 O 11 / 342.34 g/mol = 0.0590 moles Sucrose

400 g H2O / 18.02g H2O = 22.2 moles

0.0590 / 22.259 = 0.00265

 :delta: P = (0.00265)(23.8 mmHg) = 0.0631 mmHg = 0.0000830 atm


Solutions vapor pressure (atm):

X 1 22.2 moles H2O / 22.259 = 0.997

0.997 X 23.8 mmHg = 23.7 mmHg = 0.0312 atm.


Vapor Pressure at 100 degrees C (atm):

(.997) X 760mmHg = 758 mmHg = 0.997 atm


8.89 grams of sucrose (C12H22O11) is dissolved in 34.0g of water. Calculate the boiling point of this solution.

Boiling Point of Solution:

 :delta: T b = K b x m

8.89g of Sucrose (Solute)

34.0g of H 2 O (Solvent)

34.0 g / 1000g = 0.034 kg

8.89g / 342.3 g/mol = 0/0260 moles

0.0260 moles / 0.034 kj = 0.765 m

:delta:Tb = (0.52 C/m)(0.765m) = 0.398 degrees C

Are my answers, units, significant figures correct?








Offline Rabn

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Re: Vapor Pressure and Boiling Point - Check my Work
« Reply #1 on: May 07, 2009, 03:32:41 AM »
..the boiling point of the solution in question 2 is 0.398 degrees C? I don't want to sound like a jerk, but not only aren't you answering the question, the answer you got may or might not make any sense. It should be apparent why you haven't gotten more responses.

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