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Chemistry question

BUmiken12

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    I would guess it would decrease the freezing point because of freezing point depression (colligative property). The NaCl (just like anyother solute) would hinder the cyclohexane from forming it's lattice; therefore, one would need lower temps to freeze it.
     
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    bentz

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      I don't think it would have any effect on the freezing point. The two molecules don't mix and thus there's no interaction to break the tendency for the pure molecule to solidify. I believe formula is: Tf=kf*m. if there is no dissolve solute in the solution, you can't find the molality mixture. Then, the freezing point depression is zero. anyone thinks otherwise???
       

      Tweetie_bird

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        hey guys!
        I am back from a long "vacation." (more on that later)
        I think that NaCl, although it won't dissolve in cyclohexane, would have an effect on the fp. Reason being is that freezing point is a colligative property, which relies solely on NUMBER of molecules, not TYPE of molecules in a substance. Atleast that is my impression of it.

        Tweetie
         

        kutastha

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          •••quote:•••Originally posted by Tweetie_bird:
          •hey guys!
          Reason being is that freezing point is a colligative property, which relies solely on NUMBER of molecules, not TYPE of molecules in a substance. Atleast that is my impression of it.•••••Yet there are no molecules in the solution. Take a shot glass, put it in a pot of water and and see what temperature the water boils.
           

          DarkChild

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            Tweetie - I think melting point elevation/ boiling point depression - hinges on the fact that the introduced compound dissolves into the liquid. Since NaCl isnt going to dissolve in cyclohexane - think about it as dropping a rock into water - does that change the boiling point?
             

            Tobtolip

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              I have to agree with Tweetie Bird, isnt freezing point depression a colligative property?? Which means it depends solely on the number of molecules, not the type etc. Even if NaCl doesnt dissassocate its still a molecule, and your adding more of it;) So I think the freezing point would drop
               

              Hallm_7

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                Freezing point depression is a colligative property. A colligative property refers to substances that are IN SOLUTION. Since the solvent is nonpolar, NaCl will not readily dissociate in it and is therefore not in solution. Colligative properties do not apply in this case. There may be VERY insignificant FP depression due to random NaCl molecules floating through the solvent, but it is so miniscule that there is no need to calculate it.
                 

                Sm00th13

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                  In freezing-point depression, if the solute particles interfere with the process of crystal formation the occurs during freezing, the solute particles lower the temperature. There would be some weak type of intermolecular forces between the salt and cyclohexane and decrease fp. Look at it this way, the salt would take up space (b/c the salt would not dissolve and would be larger as opposed as separate ions) where cyclohexane would be, causing some disturbance in crystal formation. Now that I'm reading the previous, collgative properties does deal with solution. Solution is a homogeneous (same phase) as oppose to heterogeneous, which would be the Nacl and cyclohexane. So I agree with the previous post that fp wouldn't change much. :cool: :cool:
                   

                  Tobtolip

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                    Hmm... I now have a new understanding of colligative properties (just when you thought you had general chem down solid)... I say keep more of these questions coming =P Its good MCAT practice I think =P
                     

                    Epi

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                      no effect on fp. molecules have to be dissolved in the solution for it to effect fp. think of it on a molecular level, you cool it down, cyclohexane molecules in solution lose energy and want to form an orderly compact solid. If there is nothing in solution to hinder this movement into a solid, then fp will not change.
                       
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