Question

Part A In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 °C. If 3.10 g of CaCl2 is added to the calorimeter what will be the final temperature of the solution in the calorimeter? The heat of solution AHoln of CaCla is -82.8 kJ/mol Assume that the specific heat of the solution formed in the calorimeter is the same as that for pure water. C, 4.184 J/g C Express your answer with the appropriate units. View Available Hint(s) ? HA Value Units Submit

Answer 1

Giveni Ti = 23°c meadly = 3.10g onsoln = -82.8 kJ lmol Varo = loomL we know, Molarmass of cache=116.9849 lmoloc ando un on solna - 82.8kJ/mol 1 This is heat perlmole but actual moles of cach is not No. of moles of cach = wt = 3.109 - molar mass 110.9248 lmol No. of moles of each = 0.027 moles q = Heat per 0.027 moles of cach a 2 - 82.8kJ X 0.027mol =-2.23kJ met q=-2.23 x103 J qv = -2 230 J) q = maat m = mass of solution = mass of water t mass of eact

Density = mass volume Density of water Volume x density mass - mass = wollt 19 HAL mia 90 = 1009 Total mass i mads of 120+ mass of caca loog + 3.log - = Total mass 103-10g q=mcoT - 22305 = (103-109) (4-130 ) ( Tp-23'c) -23305 =(431.379) (Tę – 22°c).. - 22308 431-376 = Tf -23°c - 5.16c = Tf -28°C. Tf = -5-16C + 28°C If = 17.840