Question

In a coffee-cup calorimeter, 11.0-g sample of solid CaCℓ2 is dissolved in 125 g of water at 25.0 oC. The temperature in the calorimeter is measured to be 39.2 oC when the dissolution of CaCℓ2 is completed. Assuming that the specific heat of solution is equal to that of water, i.e., 4.184 J/g oC, calculate the heat of solution of CaCℓ2 in water, in kJ/mol.

Answer 1

Given for solution:

m = 125 g + 11.0 g = 136 g

C = 4.184 J/g.oC

Ti = 25 oC

Tf = 39.2 oC

use:

Q = m*C*(Tf-Ti)

Q = 136.0*4.184*(39.2-25.0)

Q = 8.08*10^3 J

= 8.08 KJ

This heat is supplied by 11.0 g of CaCl2

Molar mass of CaCl2,

MM = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

mass(CaCl2)= 11.0 g

use:

number of mol of CaCl2,

n = mass of CaCl2/molar mass of CaCl2

=(11 g)/(1.11*10^2 g/mol)

= 9.912*10^-2 mol

Use:

ΔH dissolution = -Q / number of mol

= -8.08 KJ / 9.912*10^-2 mol

= -81.5 KJ/mol

Answer: -81.5 KJ/mol