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Inorganic : Vapor Pressure
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 Message 1 of 2 in Discussion 
From: MSN NicknameZurconium2  (Original Message)Sent: 3/13/2006 10:21 PM
Steve,
 
Hello.  How would you go about solving this problem?
 
Antifreeze is a 50/50 misture by volume of ethylene glycol and water.  The density of ethylene glycol, HOCH2H2OH, is 1.11 g/ml and the density of water is 0.998 g/ml.  Assuming that the ethylene glycol is non-volatile, calculate the vapor pressure lowering of the solution at 100. ºC, the normal boiling point of water.
 
Thanks!


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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 3/14/2006 1:41 AM
The usual formula for calculating the vapor pressure of a solution is

(1)    Psolution   =   Xsolventsolvent

where

Psolution is the vapor pressure of the solution,
Xsolvent is the mole fraction of the solvent, and
solvent is the vapor pressure of the pure solvent.

This can be rearranged to give the vapor pressure lowering, DP, which is P°solvent  -  Psolution :

(2)    DP   =   Xsolutesolvent
 
You can use either formula.  In the second, the mole fraction of solute particles is

(3)    Xsolute     =                     moles of solute particles               
                               moles of solute particles + moles of solvent

The solute (nonvolatile) is ethylene glycol, a molecular substance that gives only one "particle" when it is in solution.  The solvent (volatile) is water.  You need obtain the moles of each, starting with some arbitrary amount of solution, such as 100 mL.

Since this is a "50/50" mixture by volume, in 100 mL we will have 50 mL of ethylene glycol and 50 mL of water.  Use the given densities to convert each volume to grams, and then convert to moles by dividing each by the respective molecular weight.  Now you can now calculate Xsolute from equation (3).

P° is the vapor pressure of pure water at 100°C, which is one atmosphere, or 760 torr.  You can use either pressure unit.

Now just plug into equation (2) to calculate DP.
 

Steve