Question

12.37 grams of NH₄NO₃ are dissolved in 400 mL of 25.5 ^oC water.
The ΔH_solution of ammonium nitrate is +25.69 kJ/mole.
The molar mass of ammonium nitrate is 80.05 g/mole.
What will the final temperature (celcius) of the solution be, when the salt has dissolved completely?
Assume that the heat capacity of the solution is 4.184 J/g^oC and that the density of water is 1g/mL.

Answer 1

MW of NH4NO3 = 80.05 g/mol

calculate moles of NH4NO3

mol = mass/MW = 12.37/80.05 = 0.15454 mol of NH4NO3

so...

Qsolution = HRxn*n = 0.15454*25.69 = 3.9701326 kJ

note that...

this is endothermic, so heat is absorbed from surroundings

so

Q water = 3.9701326 kJ

Qwter = 3.9701326 *!0^3

Qwater = -m*C*(Tf-Ti)

so

mass of water = 400 mL --> 400 g

-3.9701326 *10^3 = 400*4.184*(Tf-25.5)

solve for Tf

T = -(3.9701326 *10^3 )/( 400*4.184) + 25.5

Tf = 23.1277 °C, make sense since heat is being absorbd, so it must be decreased