The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol.

Question

If 4.01 g KOH is dissolved in enough water to make a 150.0 mL solution, what is the change in temperature (°C) of the solution? (The specific heat capacity of the solution is 4.184 J/g･°C and the density of the solution is 1.02 g/mL).

Answer 1

4.01 g KOH x 1 mol KOH / 56.1 g = 0.0715 mols KOH

heat generated = 0.0715 mols KOH x 57.6 kJ/mol = 4.12 kJ

Now we can use q = mC∆T to find the change in temperature (∆T)

q = heat = 4.12 kJ = 4120 J

m = mass = 150.0 ml x 1.02 g/ml = 153 g

C = specific heat = 4.184 J/gº

∆T = change in temperature

Solving for ∆T, we have...

∆T = q / (m)(C) = 4120 J / (153 g)(4.184 J/gº)

∆T = 6.44º