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

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.

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.

## Comments

Assignment Expert23.02.18, 17:30Dear Monica Blanco,

please use panel for submitting new questions

Monica Blanco21.02.18, 21:59Calculate 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 Expert04.01.17, 13:51Dear Jake

Questions in this section are answered for free. We work on all questions

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Jake30.12.16, 18:58It should be exothermic

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