Chem I/Lab: SC1 SP18 (Kristin Rowan) entProblemlD-99440412 18 of 20 /> Part A In

Question

Chem I/Lab: SC1 SP18 (Kristin Rowan) entProblemlD-99440412 18 of 20 /> Part A In the following experiment, a coffee-cup calorimeter containing 100 mL of H20 is used. The initial temperature of the calorimeter is 23.0 C If 560 g of CaCh is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution ??-In of CIL 82.8 k.J/mol. Express your answer with the appropriate units > View Avail?ble Hint(s) Value Units Submit Next > Provide Feedback

Explanation / Answer

AjitKumaranswered this

Moles of CaCl2 used in the calorimeter = mass / molar mass = 5.60 g / 111.0 g/mol = 0.05045 mol

Given the heat of solution of CaCl2, ?Hsoln = - 82.8 kJ/mol (-Ve sign indicates exothermic reaction).

Hence heat released due to hydration of 5.60 g of CaCl2 = 0.05045 mol x (-82.8 kJ/mol) = 4.1773 kJ

Now 4.1773 kJ(=4177( J) of heat will be used to increase the temperature of calorimeter.

Hence Q = 4177 J = mxSxdT

Here m = Vxd = 100 mL x 1.0 g/mL = 100 g

S = 4.186 J/g.DegC

=> 4177 J = 100 g x ( 4.186 J/g.DegC) x (T - 23.0) DegC

=>  (T - 23.0) = 9.978

=> T = 23.0 + 9.978 = 32.978 DegC (answer)

Hence the final temperature of the solution in the calorimeter is 32.978 DegC

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