Consider the dissolution of 1.50 g of salt XY in 75.0 mL of water within a calor
ID: 537284 • Letter: C
Question
Consider the dissolution of 1.50 g of salt XY in 75.0 mL of water within a calorimeter. The temperature of the water decreased by 0.93 degree C. The heat capacity of the calorimeter is 42.2 J/degree C. The density of water (and the solution) is 1.00 g/mL. The specific heat capacity of the solution is 4.184 J/g middot degree C. a. Identify the system: write a reaction that describes the event occurring in the system. b. which of the following best describes the surroundings? Explain your choice. the water alone the solution alone (dissolved salt and water) water plus the calorimeter the solution plus calorimeter c. Determine if the reaction was endothermic or exothermic. Explain. d. Calculate the quantity of heat gained or lost by the surroundings. Keep in mind your answers to items b and c above. e. Calculate the enthalpy change, Delta H_sys, for dissolving this salt. Report the value as the total energy absorbed or released when the entire sample of salt dissolves and then on a per gram basis. What additional information would be needed to report the enthalpy change on a per mole basis?Explanation / Answer
a. XY(s) + H2O(l) ---> X^+(aq) + Y^-(aq)
b . surroundings means, the solution plus calorimeter.
c. as temperature is decreased, heat is absorbed in the process. so taht
the process is endothermic.
d.
heat absorbed(or)released(Q) = m*s*DT + C*DT
m = mass of the reaction mixture = salt + water
salt = 1.5 g
water = v*d = 75*1 = 75 g
m = 76.5 g
DT = change in temperature= 0.93 C
C = heat capacity of calorimeter = 42.2 j/c
Q = 76.5*4.184*0.93 + 42.2*0.93 = 336.92 joule.
e. DHsys = Q/m = 336.92/1.5 = 224.6 j/g
additionally molecular weight of salt(XY) is required.