Question

Dissolving 5.57 g of CaCl₂ in enough water to make 288 mL of solution causes the temperature of the solution to increase by 3.77 ^oC. Assume the specific heat of the solution and density of the solution are the same as water′s (about 4.18 J/g^oC and 1.00 g/cm³, respectively) Calculate ΔH per mole of CaCl₂ (in kJ) for the reaction under the above conditions.

Hint given in feedback

Aside, the ΔH per mole for dilution depends on the process. For example, more energy is released when starting with a large volume of water (infinite dilution), than when starting with a small volume of water. Why do you think this happens?

Answer 1

Delta H = -90.427 KJ/mol

mass of ar water = density xvolume 3 1.00g/m2 x 288 ml/cm3 mass = 288 g. [iml =lem?] Heat gained by water. q = m AT q = 288 X 4.28 X q = 45 38.4768 J 3.77 . Heat release by water 5.57 g Calla = 4538-4768J So, 1 g Calle release - 4538.4768 J 5:57 110.989 Calle release - 45 38.4768 X 110:98_ 5.57 5 90427.317 J ar go.427 KJ 1 mole (110.98 g) Calle release go 427 kJ heat. so, AH = - 90.427 kJ/mol of Calla. Negative sign reaction. indicate that heat is released by