chemistry help!!!!!!!!!!!! enthalpy

Question

3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 19.21°C. Calculate ∆H for the dissolution of the unknown solid. (The specific heat of the solution is 4.18 J/g･°C and the density of the solution is 1.20 g/mL).

Answer 1

q = ∆H = mC∆T

q = ∆H = heat = ?

m = mass = 150.0 ml x 1.20 g/ml = 180. g

C = specific heat = 4.18 J/g/deg

∆T = change in temperature = 19.21 degrees

∆H = (180. g)(4.18 J/g/deg)(19.21 deg) = 14454 J (this is ∆H for dissolving 3.15 moles of unknown)

To find molar ∆H, we divided by the number of moles...

∆H = 14454 J / 3.15 moles = 4589 J / mole = 4.59 kJ/mole (to 3 signifiant figures)