enthalpy change

Question

When 3.02 g of NH₄Cl  is dissolved in enough water to make 20.05 mL of solution, the temperature dropped from 19.8°C to 9.1°C. Calculate the enthalpy change (in kJ) when 1 mole of NH₄Cl is dissolved in water. The density and specific heat of water are 1.00 g/mL and 4.18 J/g°C respectively.  

Answer 1

q = mC∆T

q = heat

m = mass = 20.05 ml x 1g/ml = 20.05 g + 3.02 g = 23.07 g (note: in some intro courses, they use only the mass of water, i.e. 20.05 g, but in more advanced courses they will add the mass of solute).

C = sp. heat = 4.18 J/g/deg

∆T = change in temperature = 19.8 - 9.1 = 10.7 degrees (note: this is an endothermic reaction)

Solving for q:

q = (23.07 g)(4.18 J/g/deg)(10.7 deg) = 1032 J

This is the heat for dissolution of 3.02 g NH₄Cl. The question asks for enthalpy change for 1 mole NH₄Cl.

3.02 g x 1mole/53.49 g = 0.0565 moles

∆H per mole = 1032 J/0.0565 moles = 18,265 J/mole = 18.3 kJ/mole