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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: DMOC on December 13, 2009, 11:17:15 AM

Title: Finding vapor pressure and freezing point depression
Post by: DMOC 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.

Title: Re: Finding vapor pressure and freezing point depression
Post by: stewie griffin 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.

Title: Re: Finding vapor pressure and freezing point depression
Post by: DMOC 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

Title: Re: Finding vapor pressure and freezing point depression
Post by: savy2020 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.