In the laboratory a general chemistry student finds that when 1.48 g of CaCl2(s)

Question

In the laboratory a general chemistry student finds that when 1.48 g of CaCl2(s) are dissolved in 112.00 g of water, the temperature of the solution increases from 22.34 to 24.71 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.53 J/°C. Based on the student's observation, calculate the enthalpy of dissolution of CaCl2(s) in kJ/mol. Assume the specific heat of the solution is equal to the specific heat of water. ?Hdissolution = kJ/mol

Explanation / Answer

Ans. Step 1: Heat change for the solution/sample is given by-

q = m s dT                            - equation 1

Where,

q = heat change

m = mass of sample

s1 = specific heat of sample = 4.18 J g-10C-1

dT = Final temperature – Initial temperature

# Total mass of sample = 1.48 g + 112.00 g = 113.48 g

Putting the values in equation 1-

            q = 113.48 g x 4.18 J g-10C-1 x (24.71 – 22.34)0C = 1124.20 J

# Step 2: Heat gained by calorimeter is given by-

            q2 = Ccal x dT                                   - equation 2

                                    - where, Ccal = Calorimeter constant

            So, q2 = 1.53 J 0C-1 x 2.370C = 3.6261 J

Note: Please RECHECK if Ccal = 1.53 J /0C (seems to be practically very low ). If its’ the same, its ok. If its different, recalculate the values as explained.

# Step 3: Total heat absorbed by the solution and calorimeter during increase in temperature must be equal to the total heat released during dissolution of CaCl2­.

So,

            Total heat released during dissolution of CaCl2, q3 = - (q + q2)

                                                                        = - (1124.20 J + 3.6261 J)

                                                                        = -1127.8261 J

Note: The –ve sign of q3 indicates release of heat during dissolution of CaCl2.

# Step 4: Moles of CaCl2 = Mass/ MW = 1.48 g/ (110.9834 g/ mol) = 0.01334 mol

Now,

            dHdis = Total heat released during dissolution / Moles of CaCl2

                        = (-1127.8261 J) / 0.01334 mol

                        = -84544.6851 J/mol

                        = -84.54 kJ/mol

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