Answer to Question #17063 in Organic Chemistry for Kenton

Question #17063
(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?
Expert's answer
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
This process is endothermic.

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Assignment Expert
23.02.18, 17:30

Dear Monica Blanco, please use panel for submitting new questions

Monica Blanco
21.02.18, 21:59

Calculate the minimum value of ∆S (J mol-1 K1) associated with dissolving NaNO3 in water at 23oC. For this process ∆H=25.41 kJ mol-1.

Assignment Expert
04.01.17, 13:51

Dear Jake Questions in this section are answered for free. We work on all questions and publish answers after verification. There is no guarantee of answering certain question but we are doing our best. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.

30.12.16, 18:58

It should be exothermic

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