Question

5.17 grams of CaCl2 was added to 85.0 cm3 of water in a Styrofoam cup calorimeter. The temperature of the water changed from 21.4°C to 27.5 °C.
Calculate qrxn for dissolving 5.17 grams of CaCl2 in 85.0 cm3 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).

Answer 1

$$Q = cm\Delta t$$ $$m = m_{water} + m_{CaCl_2}$$ $$m_{water} = d_{water}V = 1.0(g/cm^3)*85.0(cm^3) = 85g$$ $$Q = 4.18 J*g^{-1}*^oC^{-1}*(85.00+5.17)g*(27.5-21.4)^oC = 2 299.15 J*g^{-1} \approx 2300 J$$

$$\Delta H = q_{rxn} = -q_{cal} = -Q$$ $$\Delta H = -2300J$$

In J/mol:

$$\Delta H_{molarity} = \Delta H/n_{CaCl_2} = \Delta H/(m_{CaCl_2}/M_{CaCl_2}) = \Delta H *M_{CaCl_2}/m_{CaCl_2}$$

$$\Delta H_{molarity} = -2300 J*110.98g*mol^{-1}/ 5.17 g = -49372.15 J/mol \approx -49.37 kJ/mol$$

Answer: -49.37 kJ/mol