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 Message 1 of 3 in Discussion 
From: MSN NicknameIron-soldier1  (Original Message)Sent: 4/27/2008 2:08 AM
Which of the following solutions will have the lowest freezing point , 0.010m Rb I, 0.010 m K2SeO4, 0.035 m CH3OH, 0.015 m SrBr?


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 Message 2 of 3 in Discussion 
From: MSN Nickname·Steve·Sent: 4/27/2008 7:23 AM
The solution with the lowest freezing point will have the highest concentration of solute "particles."  For ionic solute, the particles are the positive and negative ions.
 
For example, the first solution, 0.01 m RbI, is 0.01 m in Rb+ ion and 0.01 m in I�?/SUP> ion, for a total solute particle concentration of 0.01 m  X  2  =  0.02 m.  This is ignoring any ion pairing in the solution, which results when an Rb+ ion comes in temporary contact with an I�?/SUP> ion.  An ion pair such as this behaves as one particle rather than two.  Since some ion pairing will occur, the actual solute particle concentration is something less than the ideal value of 0.02 m.  This is the basis of the van't Hoff factors for ionic solutes.  The van't Hoff factor for 0.01 m RbI may be about, say, 1.9 instead of the ideal 2, for example.
 
For molecular solutes like CH3OH, there is only one solute particle, the molecule itself.  Therefore, the total solute particle concentration of 0.035 m CH3OH is 0.035 m.
 
 
Steve
 
 

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 Message 3 of 3 in Discussion 
From: MSN NicknameIron-soldier1Sent: 4/27/2008 4:17 PM
Thank you for your explanation. Now I have a better concept.