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Topic: Physical Properties of Solutions  (Read 4655 times)

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Offline psychfan29

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Physical Properties of Solutions
« on: January 30, 2008, 09:00:09 PM »
1.  Can An aqueous solution be prepared that is 30% by mass of KNO3 at 20 degrees Celsius?  Explain.
I can't make heads or tails from my book.  How can you tell???

2.  Are there any exceptions to the general rule that a supersaturated solution can be made to deposit excess solute by cooling?
Again, I can't find this anywhere in my textbook, so can somebody please explain to me how you can tell?

3.  An unknown compound consists of 33.81% C, 1.42% H, 45.05% O, and 19.72% N by mass.  A 1.505-g sample of the compound, when dissolved in 50.00 mL benzene (d=0.879g/mL), lowers the freezing point of the benzene to 4.79 degrees Celsius.  What is the molecular formula of the compound?
Here's what I have so far, and I'm not even sure if this is right:
Change in freezing point= 5.53-4.7=0.83 degrees Celsius
0.83/5.12= 0.162 molal
I don't know where to go from here. I thought you would multiply by kg of solution, but that isn't given.  I tried using L of solution, but that definitely was not right. *delete me*  :o


Offline Borek

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Re: Physical Properties of Solutions
« Reply #1 on: January 31, 2008, 03:04:37 AM »
1.  Can An aqueous solution be prepared that is 30% by mass of KNO3 at 20 degrees Celsius?  Explain.
I can't make heads or tails from my book.  How can you tell???

I would start looking for KNO3 solubility in water.
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Offline ARGOS++

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Re: Physical Properties of Solutions
« Reply #2 on: January 31, 2008, 08:40:48 AM »
Dear Psychfan29;

To Q. 2:
You may think about that generally two “Things” are required for crystallisation out of a solution; a saturated or supersaturated solution and what we call “Kristallkeime” (German) and I would translate with “Seed Crystals”! Therein it is not required that “Seed Crystals” must be of the same mater as the later crystals.

To Q. 3:
Remember:   Mola Lity = Moles per Kilogram Solvent (- Not Solution! -).
So you have two concentration answers you can combine and rearrange the equation to isolate 
for 1.0 mole (= MW):
  • A.)        0.162 molal = 0.162 mole / 1000g = 1.505g / (50mL * 0.879g /mL)
    ===>  1.0 mole (= MW) =  1.505g / (50mL * 0.879g /mL) * 1000g / 0.162 mole  = y g /mole.
  • B.)   Now use the mass% to determine the indices for the molecular formula.
    You will end in a multi nitrified product you can search for with the molecular formula on:

I hope to have been of help to you.

Good Luck!
                    ARGOS++

« Last Edit: January 31, 2008, 08:54:25 AM by ARGOS++ »

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