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Topic: Finding vapor pressure and freezing point depression  (Read 13091 times)

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Offline DMOC

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Finding vapor pressure and freezing point depression
« on: December 13, 2009, 11:17:15 AM »
0.010 m Na3PO4 in water
0.020 m CaBr2 in water
0.020 m KCl in water
0.020 m HF in water (HF is a weak acid.)

(a) Assuming complete dissociation of the soluble salts, which solution(s) would have the same boiling point as 0.040 m C6H12O6 in water? (C6H12O6 is a nonelectrolyte.)

(b) Which solution has the highest vapor pressure at 28 degrees C?

(c) Which solution would have the largest freezing-point depression?

I was able to solve A quite easily (Na3PO4 and KCL) but I am confused on parts B and C. My notes don't seem to have a clear way to find the vapor pressure. Also I'm confused with C because while I know that the equation for C is just (temperature = K-constant * molality), when I use that it seems like all the solutions with 0.020 molality have the largest freezing point depression.

The answer for B is HF and C is CaBr2 so I just have to understand the process of getting there.

Offline stewie griffin

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Re: Finding vapor pressure and freezing point depression
« Reply #1 on: December 13, 2009, 11:50:56 AM »
I believe you are correct for part A.
For part B I don't remember b/c I haven't done these calculations in like 10 years.  :(
For part C I think that you're forgetting that this is a colligative property. Your equation needs to take into account the fact that when you dissolve a mole of KCl in water you get two moles of ions. However, when you dissolve one mole of CaBr2 you get three moles of ions. Thus even though they both have a concentration of .02 m to start with, one substance (the CaBr2) produces more ions in solution and it is these ions that result in freezing point depression. Take a look at your notes again and see where you account for the number of ions produced and hopefully it'll make more sense.

Offline DMOC

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Re: Finding vapor pressure and freezing point depression
« Reply #2 on: December 13, 2009, 12:07:53 PM »
AH, yes I understand now.

So it's CaBr2 with largest freezing point depression because it's basically 0.02 times 3, whereas KCl and HF are both 0.02 times 2 and Na3PO4 is .01 times 4

Now I Just need to figure out part B. Thanks!  ;D

Offline savy2020

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Re: Finding vapor pressure and freezing point depression
« Reply #3 on: December 13, 2009, 02:02:16 PM »
For part B, You don't need to find out the Vapor pressure, Relative lowering of vapour pressure is a colligative property. So proceed similarly as C and find which has smallest Lowering of vapour pressure.
:-) SKS

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