In a calorimetry experiment 8.25 g of CaCl 2 ( MM = 110.98 g/mol ) at 25.00 o C
ID: 1045591 • Letter: I
Question
In a calorimetry experiment 8.25 g of CaCl2 (MM = 110.98 g/mol ) at 25.00 oC was dissolved in 102 g of water at the same temperature. After a few minutes, the temperature of the solution rose to 35.50 oC. (Assume the specific heat capacity of the solution (mostly water) = 4.184 J/g°C and that ALL the heat from the dissolution stays in the solution formed & none is transferred to the calorimeter.) Calculate the following for this solution.
{ NOTE: Give all answers for the following in standard form. Do NOT use scientific notation. Also remember to use the rules of SFs in giving final answers. }
i) ?Hsoln = ___ J
ii) moles of CaCl2 = ___ mol
iii) molar ?Hsoln = ___ kJ/mol CaCl2
Explanation / Answer
mass of CaCl2 = 8.25 g
mass of water = 102 g
mass of solution = 102 + 8.25 = 110.25 g
specific heat = 4.184 J / goC
temperature rise = 35.5 - 25 = 10.5 oC
Q = m Cp dT
= 110.25 x 4.184 x 10.5
= 4843.5
i) Hsoln = 4843.5 J
ii)
moles of CaCl2 = 8.25 / 110.984 = 0.0743
moles of CaCl2 = 0.0743 mol
iii)
delta Hsolution = Q / n
= - 4843.5 x 10^-3 / 0.0743
delta Hsolution = - 65.2 kJ/mol