[sweetdaisy186]

I don't think I understand the question. Any help on how to start the problem would be great! Thanks!!!

When phenol, C₆H₅OH, is dissolved in bromoform, CHBr₃, it is partially associated into dimers (double molecules). The freezing point depression in a solution is 2.58 g phenol in 100.0 g bromoform is 2.374 degrees C. The freezing point of bromoform is 8.1 degrees C and K_f = 14.1 degrees Cm^-1. What fraction of the phenol is present as dimers?

[savoy7]

Some hints:

1)  Change T_f = molality x dissociation factor x K_f

2)  molality:  they give you what you need in the problem to determine it

3)  change in temperature - stated in the problem

4)  K_f for bromoform is given in the problem

You should be able to solve for the dissociation factor.  A hint on that:  If it was 1, the solute would stay together as one particle.  If it was 2, the solute would dissociate into 2 particles.

With the above information, one can figure out the answer.  Give it a try.  Post your work and someone will get back to you.

[sweetdaisy186]

Thanks! I hope I got this right! I have attached my work, I hope it shows up okay!

[xiankai]

umm... it seems too big for comfort  

[savoy7]

Looking at what you did -

If 0% of the phenol dimerized, you got 0.168 m for the solution.  That's looks good.  Meaning - I looked at it and it made sense.

What would happen if 100% of the phenols dimerized, what would the molarity have to be to produce a depression of 2.374 degrees C?

[sweetdaisy186]

Oops! Sorry that it got so big! How can I change the size? Hmm, about your question Savoy, would it be .137 M if all the dimers were dissolved?