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March 25, 2017

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In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23 degrees Celsius. If 4.00 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution of CaCl2 is -82.8 kJ/mol.

  • chemistry - ,

    Here are my numbers:
    100mL water, Temp 1: 23C, Temp2: ?, 7.20 g CaCl2, DeltaH=-82.8kJ/mol.

    First:
    Temperature is going to increase, so turn your -82.8kJ/mol to +82.8.
    Also, 100 mL water = 100 g water b/c water's density is 1g/mL

    Second:
    82.8kJ/mol=82,800J/mol

    Third:
    Molar mass of CaCl2=110.9 g/mol

    Fourth:
    (82,800 J/mol)(1mol/110.98g)=746.08 J/g

    Fifth:
    746.08J/g * 7.20 g = 5371.78 J

    Sixth:
    (100g + 7.2 g total mass)(4.18 specific heat)(T2-23C)=5371.78J

    T2-23=12
    T2=35C

    HOpe this helps!

  • chemistry - ,

    Wow finally someone who actually was able to help me get the right answer!!! ^ (I'm not sure where you got 7.2g for CaCl2, but apart from that you did everything right!)

    Thanks for the help!

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