Question

Calorimetry 1. Consider the following data: 4.99 g g 50 g 24 C 38.7 °C 111.1 g/mol Mass of CaCl Mass of Water Intial Temperature Final Temeprature Molar Mass of CaCl2 (a) Calculate the heat of solution. (AH) Assume that the specific heat of the solution is the same as water, 4.184 J/g- C and that no heat is gained or lost by the cup. (4 pts) (b) Calculate the number of moles of CaCl2. (3 pts) (c) Calculate the AH/mol (in kJ/mol) for the reaction. (3 pts) b 2. If 40 mL of 2.50 M of HCl is mixed with 40 mL of 2.50 M NaOH both at 22.9 C (a) Calculate the heat released when these two solutions are mixed if the temperature increases to 39.6 °C.(Neglect any heat absorbed by the calorimeter, assume 40 mL = 40 g and that the specific heat of the solution is 4.184 J/g-°C) (4 pts) HOY (b) Calculate the number of moles of HCl used. (3 pts) DR (c) Calculate the heat released per mole of HCI (in kJ/mol). (3 pts)

Answer 1

#1: (a): Mass of solution, m = 50g + 4.99g = 54.99 g

specific heat of solution, s = 4.184 J/g.^oC

Change in temperature, $\Delta$T = Tf - Ti = 38.7 ^oC - 24 ^oC = 14.7 ^oC  

Since temperature is increased, heat is absorbed by water and heat is released due to the formation of solution.

Heat absorbed by water, q_w = m*s*$\Delta$T = 54.99 g * 4.184 J/g.^oC * 14.7 ^oC  

=> q_w = 3382 J

=> Heat of solution, $\Delta$H = - q_w = - 3382 J or - 3.382 kJ (Answer)

(b): Number of moles of CaCl2 = mass/molar mass = 4.99 g / 111.1 g/mol = 0.0449 mol (Answer)

(c): $\Delta$H / mol = - 3.382 kJ / 0.0449 mol = 75.3 kJ/mol (Answer)