Answer to Question #160060 in Chemistry for Arvin Shopon

Question

In the calorimeter, the student dissolved 4.25 g of solid ammonium nitrate in 60.0 ml of water with an initial temperature of 22.0 °C. When the ammonium nitrate was completely dissolved, the temperature of the solution was 16.9 °C. Based on the measurement results, calculate the heat of dissolution of ammonium nitrate in kJ/mol.

Answer 1

According to the equations:

ΔH_soln = -q / n

where ΔH_soln - molar enthalpy (heat) of a solution, q - amount of energy (heat) released by water, n = moles of solute.

n = m / M

where n - moles of solute, m - mass of solute, M - molar mass of the solute.

q = m × C_g × ΔT

where q - amount of energy released by water, m - mass, C_g - specific heat capacity, ΔT - change in temperature.

From here:

ΔH_soln = - q / n = - (m_water × C_g × ΔT) / n = - (m_water × C_g × ΔT × Mr) / m = - (60 g × 4.18 J/g°C × (16.9 °C - 22 °C) × 80 g/mol) / 4.25 g = 24077 J/mol = 24.08 kJ/mol

Answer: 24.08 kJ/mol