Explanation:
First, calculate the mass of water:
m_water = 50.0 g
Next, calculate the temperature change (ΔT):
ΔT = T_final - T_initial = 18.4 °C - 23.0 °C = -4.6 °C
Since the temperature decreases, the process is endothermic (heat is absorbed). The heat absorbed (q) is:
q = mcΔT
where m is the mass of water, c is the specific heat capacity of water (approximately 4.184 J/g°C), and ΔT is the temperature change:
q = 50.0 g × 4.184 J/g°C × (-4.6 °C) = -966.4 J
Since the process is endothermic, the enthalpy change (ΔH) is positive:
ΔH = q = 966.4 J
Next, calculate the number of moles of ammonium nitrate (n):
n = m / M
where m is the mass of ammonium nitrate (2.88 g) and M is the molar mass (80.06 g/mol):
n = 2.88 g / 80.06 g/mol = 0.036 mol
Finally, calculate the molar enthalpy of solvation (ΔsolvHm):
ΔsolvHm = ΔH / n = 966.4 J / 0.036 mol = 26,822 J/mol = 26.82 kJ/mol
Therefore, the molar enthalpy of solvation (ΔsolvHm) is 26.82 kJ/mol.
