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Posted by on Monday, February 20, 2012 at 4:23pm.

I had to perform an experiment for the freezing point depression in which we had to determine the frezzing points of water as a pure solvent, sodium chloride, sucrose, and ethylene glycol.
We had to work with a partner so I completed the experiment for water, NaCl, and the glycol and had to get the info for sucrose from my partner. The data that was collected was the freezing pint of DI water which was the corrected value for the thermometer that I used. We had to record the temp every 30 seconds until we saw ice crystals. We had to do the same thing for the NaCl, sucrose, and gycol. For the NaCl we started at 0.5 min in which the temp was 15.0 degrees celsius and crystals formed at 10.5 min at -2.0 degrees celsius.

For my calculations the first thing I need to calculate is the delta Tf. Is this just the initial temp - the final temp?



Chemistry - Hannah, Saturday, February 18, 2012 at 4:22pm
The very frist thing that was asked for the calculations was the freezing point in degrees celsius for NaCl and Sucrose and I put -2.0C for NaCl and for sucrose my partner got -4.0C.


Chemistry - Hannah, Saturday, February 18, 2012 at 5:14pm
In my lab book it just says delta Tf = the change in the freezing point (Tsolution - Tpure solvent = negative number)


Chemistry - DrBob222, Saturday, February 18, 2012 at 6:56pm
In your initial post you said you had to determine the freezing point of pure water (that's the solvent). So delta T = f.p. soln - f.p. pure water.

I understand that to get delta T i had to do the fp of the soln - fp of water. My freezing point of the DI water was 0.5C so I have to subtract this from the fp that I got for NaCl and sucrose, is that correct?

  • Chemistry - , Monday, February 20, 2012 at 4:46pm

    Frankly, it really doesn't make any difference (to me at least) how you do it; it's just the difference without regard to sign. Technically, Kf is negative but I don't bother with that either. We know the freezing point is depressed and we know the f.p. will be below the normal f.p. I simply subtract one from the other to get the difference and drop the sign if there is one.
    So let's say f.p. is -1.0 and you started at -.5.
    f.p.-.5 = -1.0-(-.5) = -1+.5 = -.5 and delta T = .5

    Now the other way.
    -.5-f.p. = -.5-(-1.0) = -.5+1.0 = .5 and delta T = .5.

  • Chemistry - , Monday, February 20, 2012 at 6:00pm

    ok so my freezing point for NaCl was -2.0 so I can do

    -2.0-(0.5)= -2.5 degrees celsius. My freezing point for DI water was positive 0.5. So delta T would be -2.5?

  • Chemistry - , Monday, February 20, 2012 at 6:07pm

    -2 -(.5) = -1.5, not -2.5.
    If the pure DI water was +0.5 and the new f.p. was -2.0, then delta T is 2.5

  • Chemistry - , Monday, February 20, 2012 at 6:53pm

    Ok and the units I use would be degrees celsius correct?

  • Chemistry - , Monday, February 20, 2012 at 7:05pm

    Now I have to calculate the mass of solvent(kg) for both NaCl and Sucrose and I have to assume that the density of water is 1.00g/mol. I am not sure how to figure this out. I think you already answered this for me once but I cant find my original post with the answer. Im sorry.

  • Chemistry - , Monday, February 20, 2012 at 7:54pm

    mass = volume x density

  • Chemistry - , Monday, February 20, 2012 at 8:55pm

    so do I use the weight that I got for NaCl and multiply by 1.00g/mol? Or the solvent in this case is water and we used 100mL

  • Chemistry - , Monday, February 20, 2012 at 9:03pm

    For both NaCl and sucrose we had to mix it with 100ml of water so for the mass of solvent (kg) I would do
    100mL X 1.00g/mL = 100g and then to get kg I would divide by 1000 so 0.1kg. This would be the answer for sucrose as well. Is this correct?

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