# Question 56d6d

Apr 2, 2015

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 ${C}_{\text{cal" = "400 J°C⁻¹}}$.

Then ${q}_{3} = {C}_{\text{Cal"ΔT = "400 J"cancel("°C⁻¹") × (-3 cancel("°C")) = "-1200 J}}$.

The mass of the water is

100 cancel("mL") × "1.00 g"/(1 cancel("mL")) = "100 g"

The mass of the solution is

$m = \text{8.037 g + 100 g" = "108.0 g}$

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

${q}_{1} + {q}_{2} + {q}_{3} = \text{0.1004 mol" × ΔH – "1360 J" – "1200 J" = 0}$

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.