If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees Celsius. Determine the molar heat of solution of the ammonium nitrate.

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁ + q₂ = 0

n = moles of NH₄NO₃ = 8.00 g NH₄NO₃ * 1 mol NH₄NO₃/80.0 g NH₄NO₃

∴ n = 0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m = mass of solution = 1000.0 g + 8.00 g = 1008.0 g

q₂ = mcΔT = 58.0 g * 4.184 J°C⁻¹ g⁻¹ * ((20.39-21) °C) = - 2570.19 J

q₁ + q₂ = 0.100 mol * ΔHsoln - 2570.19 J = 0

ΔHsoln = + 2570.19 J / 0.100 mol = + 25702 J/mol = + 25.7 kJ/mol