Chemistry

A 1.00% by mass MgSO4(aq) solution has a freezing point of -0.192°C.
(a) Estimate the van't Hoff i factor from the data.
(b) Determine the total molality of all solute species.
(c) Calculate the percentage dissociation of MgSO4 in this solution.

Can anyone help me with this at all?

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asked by Daiyo
  1. a) delta T = i*Kf*m
    i = delta T/(Kf*m).
    Substitute and solve for i. I found 1.23 but you need to confirm that.
    (You will need to find m and a 1% solution MgSO4 means 1 g MgSO4 + 99 g water; therefore, molality = moles MgSO4 (which is 1 g/molar mass MgSO4)/kg solvent (and kg solvent is 0.099 kg). I find something like 0.0839m but you need to confirm that.

    b).......... MgSO4 ==> Mg^+2 + SO4^-2
    start ...0.0839....0........0
    change.........-x.....x.........x
    end........0.0839-x....x.........x

    total = 0.0839-x + x + x = 1.23*0.0839
    Solve for x which is the amount ionized.
    c) Then percent ion = (amount ionized/starting amount)*100 = ?? and I find something like 23 or 24%. (The starting amount is 0.0839)

    Having done all of this I want to state that I STRONGLY disagree with the idea of giving this type problem to students. It fosters the incorrect idea that MgSO4 really has a percent ionization that ISN'T 100%. I have seen similar problems posted on Jishka and in problem texts in which NaCl is treated the same way. Nonsense. NaCl is 100% ionized, MgSO4 is 100% ionized. This 23% for MgSO4 in this problem is nonsense. While it is true that the solution behaves as if it were 23% ionized, it ionizes 100% and any deviation from that is because it is not an ideal solution. Read up on the Debye-Huckel theory and activity coefficients if you want to find the real reason why it behaves this way. And while I'm at it, I should point out that there is a MUCH easier way to go about the fake percent ionization. Note that i = 1.23, and note if it were not ionized at all i = 1.00, then (1.23-1.00)*100 = guess what? 23% and without all of that goop above.

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  2. Thanks DrBob, that's what i was thinking about the ionization percentage of the MgSO4 as well but I figured out the problem with a little help from my TA today.

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    posted by Daiyo

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