Question

The aluminum cup inside your calorimeter weighs 40.85 g. You add 49.81 g of water and 3.03 g of KCl to the calorimeter. The initial temperature is 20.1oC, and the final temperature is 16.9oC. What is the heat of dissolution for the amounts of salt added, in units of J? Assume that: the calorimeter is completely insulated the heat capacity of the empty calorimeter is the heat capacity of the aluminum cup. the mass of KCl added is small enough to ignore (i.e. the mass of the solution is the mass of water) the heat capacity of the water is unchanged: 4.184 J g-1 oC-1. Perform all calculations without rounding, but then provide your answer to the correct number of significant figures.

Answer 1

Ans.

Hence, Heat of dissolution of KCl = 1.9x 10⁴ J mol^-1

Step 1: The amount of heat change for a sample is given by- Step 2: Mass of KCI 3.03 g = m SA,AT -equation 2 Where, CcalCalorimeter constant 3.030 MW 74.551 g mol equation 1 Moles of H2SoMass = 0.0406 mol Where, mmass of sample specific heat of sample Temperature at thermal equilibrium-Initial temperature Now s AT Given AT-1690 -20.10)c ater49.81 g 19309.58 J mol 3.20 °c mol of sample 0.0406 mol motal 49.81 g 1 0-1 mAI 40.85 g S 0.90 Jg C At thermal equilibrium, total heat released during dissolution is equal to the total amount of heat gained by the calorimetric system (qs for waterqAl for calorimeter). Now, qqs (heat gained by water) qAI (Heat gained by calorimeter Al) 49.8 gx 4.184, g.1°C1x-320%) +40.9gx 090, g.Pcx-3.20%) -q 666.9 J+-117.9 J 9784.81