Question

When a 3.00-g sample of KCl was added to 3.00 × 102 g of water in a coffee cup calorimeter, the
temperature decreased by 1.05 °C. How much heat is involved in the dissolution of the KCl?
please show work thank you

Answer 1

Mass of water = 3.00 times 10^2 g Specific heat (s) of water is 4.184 J/g degree C Change in temperature, Delta t = -1.05 degree C Enthalpy of the solution, q = ms delta t = (3.00 times 10^2 g) times 4.184 J/g degree C times (- 1.05 degree C) = -1.32 times 10^3 J = -1.32 KJ Mass of KCl = 3.00 g Number of moles of KCl = 3.00 g times mol/74.6 g = 0.0402 mol Calculate the enthalpy of the reaction for one mole. = -1.32 kJ/0.0402 mol = -32.7 kJ mol^- 1 Hence, - 32.7 KJ mol^- 1 heat is involved.