concerning the "i" for freezing pt depression

[orangepopsicle]

The question in DATQvault was:
Find the freezing point of a 2.0m solution of CH3OH in chloroform, CHCl3. The normal freezing point of chloroform is -63.5°C. The kf for chloroform is 4.7°C/m.

Their answer for the new freezing pt was:
(-63.5)-(4.7 x 2.0 x 1) °C
I thought for sure the i for this question would be 2 since CH3OH, wouldnt it break up into CH3O- + H+? How come in the question they have the i as 1? and if this is true, then how come CH3OH won't break up into 2 ions and will stay a whole compound, and how can you know that for sure?

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[510586]

The question in DATQvault was:
Find the freezing point of a 2.0m solution of CH3OH in chloroform, CHCl3. The normal freezing point of chloroform is -63.5°C. The kf for chloroform is 4.7°C/m.

Their answer for the new freezing pt was:
(-63.5)-(4.7 x 2.0 x 1) °C
I thought for sure the i for this question would be 2 since CH3OH, wouldnt it break up into CH3O- + H+? How come in the question they have the i as 1? and if this is true, then how come CH3OH won't break up into 2 ions and will stay a whole compound, and how can you know that for sure?

I think i is only for ionic compounds and strong electrolytes (strong acids strong bases). Although can someone clarify if COOH compounds are treated differently?

 

[orangepopsicle]

^^can someone confirm this?? and I would also like to know if COOH compounds are treated differently.

 

[orangepopsicle]

Actually I don't think the Van Hoff factor is just for strong acids and strong bases, since there are problems that list Na2SO4 which gives an i of 3 and it is not a strong acid or a base. Can someone explain if ROH's are unable to dissociate and why?

 

[510586]

Actually I don't think the Van Hoff factor is just for strong acids and strong bases, since there are problems that list Na2SO4 which gives an i of 3 and it is not a strong acid or a base. Can someone explain if ROH's are unable to dissociate and why?

That's why I said ionic compounds

 

[cacajuate]

Actually I don't think the Van Hoff factor is just for strong acids and strong bases, since there are problems that list Na2SO4 which gives an i of 3 and it is not a strong acid or a base. Can someone explain if ROH's are unable to dissociate and why?

SO4 is soluble.


Have you watched chads videos yet? Save all those practice tests till after that.

 

[orangepopsicle]

I have watched all of Chad's videos and I know SO4 is soluble. My question if you reference above was why CH3OH had a van hoff factor of 1 instead of 2, since I feel like it dissociates into CH3OH --> CH3O- + H+, and if it doesnt dissociate, and stays as a whole compound, why.

 

[510586]

I have watched all of Chad's videos and I know SO4 is soluble. My question if you reference above was why CH3OH had a van hoff factor of 1 instead of 2, since I feel like it dissociates into CH3OH --> CH3O- + H+, and if it doesnt dissociate, and stays as a whole compound, why.

It is a weak electrolyte/weak base so it's not very soluble I guess

 

[cacajuate]

I have watched all of Chad's videos and I know SO4 is soluble. My question if you reference above was why CH3OH had a van hoff factor of 1 instead of 2, since I feel like it dissociates into CH3OH --> CH3O- + H+, and if it doesnt dissociate, and stays as a whole compound, why.

Only strong acids dissociate completely. Is methanol a strong acid?

 

[orangepopsicle]

I mean isn't water a weak base as well, so does that mean H2O would be considered to have an i of 1?

 

[cacajuate]

I mean isn't water a weak base as well, so does that mean H2O would be considered to have an i of 1?

Water is what they are dissolved in.

Your making it more complicated than it is. Watch Chads video again on it, I'm guarantee you will have it then.