calorimetry....

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 11.11°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

∆H = mC∆T

∆H = ?

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

C = specific heat = 4.18 J/g/deg

∆T = change in temperature = 11.11 degrees (positive, so reaction is exothermic and ∆H will be negative)

∆H = (180 g)(4.18 J/g/deg)(11.11 deg)

∆H = -8359 Joules (this is the ∆H for the dissolution of 3.15 moles of the unknown)

If you want the molar enthalpy of dissolution, then you have...

-8359 J/3.15 moles = -2654 J/mole = -2.65 kJ/mole = molar enthalpy