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Molar Heat of Solution

  1. Mar 17, 2007 #1
    1. The problem statement, all variables and given/known data
    When 5.00 g of NaOH(s) are added to 100 g of water using a calorimeter (with Cp = 493.24 J/K), the temperature rises from 25.0 to 37.5 °C. Calculate the molar heat of solution.
    [align=center]NaOH(s)--->Na(aq)+OH(aq)[/align]
    Assume that the specific heat capacity of water is 4.18 J/gK and that of the NaOH(aq) solution is the same.

    2. Relevant equations
    1) qp = Cp∆T,
    2) | Heat Lost by Hot Water | = | Heat Gained by Cold Water | + | Heat Gained By Calorimeter |

    3. The attempt at a solution

    Heat Gained By Solution:
    (100 g H2O) x (4.18 J/gK) x (37.5 - 25)K = 5486.25 J

    Heat Gained By Calorimeter:
    Using qp = Cp∆T,
    Cp∆T = (493.24 J/K) x (37.5 - 25)K = 6165.5 J


    | Heat Lost | = 5486.25 J + 6165.5 J = 11 651.75 J = 11.65175 kJ

    Molar mass of NaOH = (22.99 + 16.00 + 1.01) = 40
    5.00 g NaOH x (1 mol ÷ 40g) = 0.125 mol

    11.65175 kJ ÷ 0.125 mol = 93.214 kJ/mol



    I was just wondering if I did that correctly. I don't actually know the correct answer, but one of my fellow students got a different answer and I'm unsure if I'm the one with the wrong answer or not.
    If this isn't the way to find ∆H, could someone show me how or perhaps point me to a useful page on the internet?

    Thanks
     
  2. jcsd
  3. Mar 13, 2008 #2
    One small thing I can see immediately

    you wrote;
    (100 g H2O) x (4.18 J/gK) x (37.5 - 25)K = 5486.25 J
    but this should actually be;
    (100 g H2O) x (4.18 J/gK) x (37.5 - 25)K = 5225 J
    or
    (105 g H2O) x (4.18 J/gK) x (37.5 - 25)K = 5486.25 J

    I don't know if this helps but it's what I could see immediately
     
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