Question

Dissolving 4.93 g of CaCl₂ in enough water to make 311 mL of solution causes the temperature of the solution to increase by 3.59 ^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?
Find the heat released per gram and convert to per mole using the MM. Is the process exothermic? If, yes, then what is the sign for ΔH?

Answer 1

volume of solution = 311 ml

denisty of solution = 1 g/ml

mass of solution = density x volume

                            = 311 g

specific heat = 4.184 J/goC

temperature increase = 3.59 oC

heat Q = m Cp dT

            = 311 x 4.18 x 3.59

           = 4667 J

moles of CaCl2 = 4.93 / 110.98 = 0.0444

H = - Q / n

       = - 4667 / 0.0444

       = 105112.6 J/mol

H = - 105.1 kJ/mol