Answer to Question #95175 in General Chemistry for Amrit Uppal

Question #95175
5.17 grams of CaCl 2 was added to 85.0 cm 3 of water in a Styrofoam cup calorimeter. The temperature of the water changed from 21.4°C to 27.5 °C.
Calculate the molar q value (q mol) for dissolving 5.17 grams of CaCl 2 in 85.0 cm 3 of water.
Assume the heat capacity of the system is due only to the water in the system. The specific heat of water is 4.18 J/(g∙°C).
1
Expert's answer
2019-09-25T05:44:35-0400

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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