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General : Vapor Pressure
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 Message 1 of 6 in Discussion 
From: MSN Nicknamegoldie647  (Original Message)Sent: 2/18/2008 12:14 AM
I am a little confused about the vapor pressure formula.

Question:
A solution is formed by dissolving 10.0g of KCL in 500.0g of H20.

What is the vapor pressure of the solution at 25 degrees Celsius if the vapor pressure of pur H2) is 23.8 mmHG at 25 degree Celsius?

So, since the question is asking for the vapor pressure of the solution I used:

Psolution = P pure solvent x Mole fraction of solvent

Is this correct?

Now for another question it asks what is the vapor pressure of a solution made by dissolving 24.6g of camphor (C10H16O) in 98.5g of Benzene. Vapor pressure of benzene is 100.0 mmHg. (Camphor is a low-volatility solid).

I tried to use the same formula :

Psolution = P pure solvent x Mole fraction of solvent

But I did not get the correct answer.

Please can you explain the formulas and what the A and B represent.

Pa = Xa x P pure a
Pb = Xb x P pure b

Thanks


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Reply
 Message 2 of 6 in Discussion 
From: MSN Nickname·Steve·Sent: 2/18/2008 5:43 PM
>>  Psolution   =   Ppure solvent   x   Mole fraction of solvent  <<

Hi, your method is correct in both cases.  In the first solution of KCl in water, the mole fraction of water, the solvent, is

XH2O    =                   moles of water               
                   moles of water  +  2(moles of KCl)

The moles of KCl are multiplied by 2 because 1 mole of KCl gives 2 moles of solute particles (1 K+ and 1 Cl�?/SUP> ion).  That will have only a small effect on your answer.
 
I'm not sure why you're not getting the correct answer for the second problem.  The mole fraction of the solvent, benzene in this case, is

Xbenzene    =                  moles of benzene                   
                       moles of benzene  +  moles of camphor

The formulas with PA and PB apply to a solution made of liquid A and liquid B.  Each liquid is volatile and will contribute a vapor pressure.  The vapor pressures of each liquid in the solution is given by those formulas.  The total vapor pressure of the solution is PA + PB.

In your problems, however, you do not have two liquids in the solution.  Instead you have a non-volatile solute dissolved in a liquid, and your method of calculating the vapor pressure of these solutions is the correct one. 

Steve

Reply
 Message 3 of 6 in Discussion 
From: MSN Nicknamegoldie647Sent: 2/18/2008 7:30 PM
Ok

Can you tell me if I am correct in making this statement:

If the solvent is volatile, such as H2O in the first problem, then you can expect the Partial Pressure of the solution to be slightly less than the Partial Pressure of the Pure solvent.

If this is true then, what about the second question with benzene as the solvent. Is benzene volatile?

Also, how can the statement "Camphor is a low-volatility solid" mean that it is non-volatile?

Thanks

Reply
 Message 4 of 6 in Discussion 
From: MSN Nickname·Steve·Sent: 2/18/2008 10:50 PM
>>  If the solvent is volatile, such as H2O in the first problem, then you can expect the Partial Pressure of the solution to be slightly less than the Partial Pressure of the Pure solvent.  <<
 
That's correct.
 
 
In the second question, benzene is the solvent, and it is volatile (its boiling point is only 80.1°C compared to 100°C for water).  Camphor is the solute which is essentially nonvolatile compared to benzene.  Actually, solid camphor does have a low vapor pressure at 25°C, 0.65 mm Hg, but I think in the problem we are supposed to assume it is nonvolatile since a vapor pressure was not given.  Plus, I doubt if the vapor pressure of camphor is the same once it is dissolved in benzene.
 
So while it is true that camphor is a "low-volatility solid", it is not truly "non-volatile."
 
 
Steve

Reply
 Message 5 of 6 in Discussion 
From: MSN Nicknamegoldie647Sent: 2/18/2008 11:55 PM
Thank you so much for clarifying this.



Reply
 Message 6 of 6 in Discussion 
From: MSN Nickname·Steve·Sent: 2/19/2008 12:38 AM
Glad to help!

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