Answer:
THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.
Explanation:
1. write out the variables given:
Mass of Calcium chloride = 13.6 g
Change in temperature = 31.75°C - 25.00°C = 6.75 °C
Density of the solution = 1.000 g/mL
Volume = 100.0 mL = 100.0 mL
Specific heat of water = 4.184 J/g °C
Mass of the water = unknown
2. calculate the mass of waterinvolved:
We must first calculate the mass of water in the bomb calorimeter
Mass = density * volume
Mass = 1.000 * 100
Mass = 0.01 g
3. calculate the quantity of heat evolved:
Next is to calculate the quantity of heat evolved from the reaction
Heat = mass * specific heat of water * change in temperature
Heat = mass of water * specific heat *change in temperature
Heat = 13.6 g * 4.184 * 6.75
Heat = 13.6 g * 4.184 J/g °C * 6.75 °C
Heat = 384.09 J
Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.
4. calculate the molar mass of CaCl2:
Next is to calculate the molar mas of CaCl2
Molar mass = ( 40 + 35.5 *2) = 111 g/mol
The number of moles of 13.6 g of CaCl2 is then:
Number of moles of CaCl2 = mass / molar mass
Number of moles = 13.6 g / 111 g/mol
Number of moles = 0.1225 mol
So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..
5. Calculate the enthalpy of solution in kJ/mol:
If 1 mole of CaCl2 is involved, the heat evolved is therefore:
Heat per mole = 384.09 J / 0.1225 mol
Heat = 3 135.43 J/mol
The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.
