First we should calculate the amount in moles of CaCl2 inserted into the water calorimeter:
M(CaCl2) = 111 g/mol;
"n(CaCl_2)=\\frac{m(CaCl_2)}{M(CaCl_2)}=\\frac{5.17g}{111g\/mol}=0.0466mol"
The temperature in the calorimeter was changed on 27.5C - 21.4C = 6.1C. The mass of water in the calorimeter is 85.0 g. Thus, the amount of heat released in the calorimeter during the dissolution of CaCl2 was equal to:
Q(CaCl2) = C(H2O)m(H2O)(t2 - t1) = 4.18 Jg-1C-1 85.0g 6.1C = 2167J
Now, it is the right time to found the specific molar heat of CaCl2 dissolution:
"Q_{spec}=\\frac{Q(CaCl_2)}{n(CaCl_2)}=\\frac{2167J}{0.0466mol}=46.5Jmol^{-1}"
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