Question

Dissolving 7.51 g of CaCl2 in enough water to make 332 mL of solution causes the temperature of the solution to increase by 3.25 oC. Assume the specific heat of the solution and density of the solution are the same as water′s (about 4.18 J/goC and 1.00 g/cm3, respectively) Calculate ΔH per mole of CaCl2 (in kJ) for the reaction under the above conditions.

Answer 1

65.72 kJ/mol

Step-by-step explanation:

The temperature change, ΔT, can be used to calculate the amount of heat absorbed by the solution:

q = CmΔT

where q is the heat absorbed, C is the specific heat capacity of water (4.18 J/goC), m is the mass of the solution, and ΔT is the temperature change.

The mass of the solution can be calculated using its density:

m = Vd

where V is the volume of the solution (332 mL = 0.332 L), and d is the density of water (1.00 g/cm3).

m = 0.332 L x 1.00 g/cm3 = 332 g

The amount of heat absorbed, q, can now be calculated:

q = CmΔT = (4.18 J/goC) x (332 g) x (3.25 oC) = 4447 J

This amount of heat is absorbed by the dissolution of 7.51 g of CaCl2. To calculate the enthalpy change per mole of CaCl2, we need to convert grams to moles:

moles of CaCl2 = 7.51 g / 110.98 g/mol = 0.0676 mol

Therefore, the enthalpy change per mole of CaCl2 is:

ΔH/mol = q / moles of CaCl2 = 4447 J / 0.0676 mol = 65720 J/mol = 65.72 kJ/mol

So the enthalpy change per mole of CaCl2 is 65.72 kJ/mol.