The molar enthalpy of solution of ammonium nitrate is +25 kJ/mol.
There are three heats to consider:
heat of solution + heat absorbed by water + heat absorbed by calorimeter = 0
#q_1 + q_2 + q_3 = 0#
#nΔH + mcΔT + C_"Cal"ΔT = 0#
I will arbitrarily assume that #C_"cal" = "400 J°C⁻¹"#.
Then #q_3 = C_"Cal"ΔT = "400 J"cancel("°C⁻¹") × (-3 cancel("°C")) = "-1200 J"#.
The mass of the water is
#100 cancel("mL") × "1.00 g"/(1 cancel("mL")) = "100 g"#
The mass of the solution is
#m = "8.037 g + 100 g" = "108.0 g"#
I assume that the specific heat capacity of the solution is the same as that of pure water.
#q_2 = mcΔT = 108.0 cancel("g") × "4.184 J"cancel("°C⁻¹g⁻¹") × (-3.0 cancel("°C")) = "-1360 J"#
#n = 8.037 cancel("g NH₄NO₃") × "1 mol NH₄NO₃"/(80.04 cancel("g NH₄NO₃")) = "0.1004 mol NH₄NO₃"#
#q_1 = nΔH = "0.1004 mol" × ΔH#
#q_1 + q_2 + q_3 = "0.1004 mol" × ΔH – "1360 J" – "1200 J" = 0"#
#ΔH = "+2560 J"/"0.1004 mol" = "+25 000 J/mol" = "+25 kJ/mol"#
Note: The answer can have only 2 significant figures, because that is all you gave for the temperature change.
