Answer to Question #17063 in Organic Chemistry for Kenton

Question

(a) When a 3.88-g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter (Figure 5.17), the temperature drops from 23.0 oC to
18.4 oc. Calculate H (in kJ/ mol NH4N03) for the solution process
NH4NO3(S) ----> NH4 +(aq) + N03 - (aq)
Assume that the specific heat of the solution is the same as that of pure water.
(b) Is this process endothermic or exothermic?

Answer 1

a)
Molar mass of NH4NO3 = 14+4+14+48 = 80 g/mol
So, 3.88 g is equal to
3.88/80 = 0.0485 mol
Now, mass of soln = 60+3.88 g = 63.88 g
Heat absorbed
= 63.88x4.18x(23-18.4) = 1228 J
So, 0.0485 mol absorbs 1228 J
1 mol
absorbs 25319 J
So, heat of dissolution of NH4NO3 is 25.319
kJ/mol
b)
This process is endothermic.