Melting Point and Impurities or Mixtures

[stester77s]

Hi, say you have two solid compounds: Compound X and Compound Y.

Compound X Melting Point: 80 degrees
Compound Y Melting Point: 120 degrees

1) If a sample of compound X is contaminated with a small amount of compound Y, the melting point will be:
a)Below 80 degrees
b) Above 80 degrees
c) impossible to predict

2) If you have a 50-50 mixture of compound X and Y, the melting point will:
a)Be less than 80 degrees
b)Be between 80 and 120 degrees
c)Be greater than 120 degrees

also please explain your choice for number 2 in particular. Thanks

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[DrknoSDN]

It's A for both I believe.
Freezing point depression occurs because the impurities obstruct the formation of a pure crystal lattice.
When the crystal lattice is disrupted it does not hold together as well so it melts more easily (at a lower temperature).


For #2 a Real World example is a 1:1 ratio of antifreeze and water.
the 50/50 mix freezes at -37 while the pure antifreeze is only -5, and water is of course 0.

 

[Czarcasm]

Don't take my word for this. It may not be factually true, but I'll try to reason it out.

I guess conceptually, it's more difficult to freeze compound X (f.p of 80) than it is to freeze compound Y (f.p 120). Adding an impurity to either of the two will depress the freezing point of pure solution - which is just a colligative property of any solution. But on a molecular level, I suppose it wouldn't make sense for compound X to freeze at a higher freezing point than it would naturally in a pure solution. That's an intensive property unique to each compound based on it's molecular structure. Therefore, it would need to reach a temperature of atleast 80 but because an impurity is added to solution, it will depress the freezing point even further and eventually both compounds will freeze (known as the eutectic point). I had a passage on this and they threw that out there, so I thought I'd share. Not a word you really need to know, but might help if you wanted to look into it more for clarity.

It might just help to think of Compound X of your solvent, and if you somehow really saturate it with a lot of solute (compound Y), it will lower the freezing point until eventually it reaches a maximal freezing point due to the introduced solute.

Bare in mind there is some maximal value. freezing point d.p = i x kf x m (there's only so many moles that can dissolve in solution per kg solvent), i is based on the amount of dissociation that occurs, and kf is some constant. Each of these values have some maximal value, therefore, freezing point will eventually depress to some maximum value. If you can somehow dissolve more of some solute into solution (before it begins precipitating), then it would increase the depression. It doesn't really matter what solute you chose, all that matters is the amount that is dissolved.

 

[Czarcasm]

My confusion though is, in a 1:1 mixture, which of the two compounds is actually considered the solute. How would one even calculate the freezing point depression?