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[Chemistry] colligative properties

  1. Oct 13, 2005 #1
    I need help on colligative properties. I'm just Really really confused so it's better to explain this to me in steps. Anyone who took chemistry or who gets it please help!!:cry: Just take a look at the problem =] thankyou

    first question What is the freezing point of a soulution of 210.0 g of glycerol, HOCH2CHOHCH2OH, dissolved in 350 g of water?

    -Formula: m=mol* # of particles/kg
    I'm trying to find m
    so far i got m=2.282608696 Mol * # of particles/.35 kg.
    I have trouble finding the # of particles.

    Second question A 0.171 g sample of an unkown organic compound is dissolved in ether. the soulution has a toatl mass of 2.470 g. The boiling point of the solution is 36.43 degree C. What is the molasr mass of the organic compound?
  2. jcsd
  3. Oct 13, 2005 #2


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    I'll start helping with the first one...

    I think what you're trying to solve for is molality, but you don't have the formula right.

    molality = moles of solute/mass of solvent

    You have the mass as a given, so you need to find the number of moles of solute (glycerol in this case). You'll have to determine the molecular weight of the glycerol (remember, the units for molecular weight are g/mol), and use that to determine the number of moles in 210.0 g.

    Do you then know how to use the molality to determine the freezing point depression?
  4. Oct 13, 2005 #3
    Hmm i'm still confused. Cuz that's the formula given by our chem teacher.
    For molarity it's =moles(# of particles/kg)
    I'm tyring to find molarity not molality l0l
    thanks tho! ^^
  5. Oct 14, 2005 #4


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    Molarity is moles/liter. Check your textbook, you may have gotten it mixed up in your notes. All the words sound very similar, so it's easy to make mistakes during notetaking.

    A mole is 6.02X10^23 molecules (that's Avogadro's number).

    Yes, you need molality, not molarity. :smile:
  6. Oct 14, 2005 #5


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    Ai : Tell us what you plan to do after you find the concentration, 'm' (which, as Moonbear says, should be a molality).

    Also, show what you've tried with the 2nd problem if you want to be helped with it.
  7. Oct 16, 2005 #6
    l0l it's ok now, i've already turned it in. About the # of particles my teacher who hardly helps at all =( said that you need to know that for the boiling pt and the freezing pt. To find it you have to look at the ion chart that we're supposed to have and see if that solution is there. If it is it tells you how much it is on the top. the [-1 or +2 that thing] if it's not there then you put 1 as the # of particles. I know how to find the # of particles but then i'm still confused is there SUPPOSE to be a # of particle for boiling pt bc he said there's one for the freezing pt.

    [question] calculate the freezing pt and boiling pt of a solution of 383 g. of glucose(C6H12O6), dissolved in 400 g of water.

    Formula that i used
    FBf=FBi+m(freezing constant)
    =0+m(freezing constant)
    Now i dunno what to do with the 400 g of water and the 383 g of glucose. Are we suppose to plug in 383 as the g to find the molarity? and where do we plug in the 400 g of water?
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