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In the laboratory a general chemistry student finds that when 2.80 g NH4Cl(s) is dissolved in 104.10 g water, the temperature of the solution drops from 25.17 to 23.28 °c. the heat capacity of the calorimeter was determined in a separate experiment to be 1.51 J/°C. Based on the student's observation, calculate the dissolution enthalpy of NH4Cl(s) in kj/mol.
Accepted Answer
The dissolution enthalpy of NH4Cl(s) can be calculated using the following steps:
1. Calculate the heat absorbed by the solution:
ΔT = 25.17 °C - 23.28 °C = 1.89 °C
q_solution = m_solution
C_solution
ΔT
Assuming the specific heat capacity of the solution is approximately the same as that of water (4.184 J/g°C), we get:
q_solution = (104.10 g + 2.80 g)
4.184 J/g°C
1.89 °C ≈ 882 J
2. Calculate the heat absorbed by the calorimeter:
q_calorimeter = C_calorimeter
ΔT
q_calorimeter = 1.51 J/°C
1.89 °C ≈ 2.85 J
3. Calculate the total heat absorbed:
q_total = q_solution + q_calorimeter ≈ 882 J + 2.85 J ≈ 885 J
4. Convert the heat absorbed to kJ:
q_total = 885 J
(1 kJ / 1000 J) ≈ 0.885 kJ
5. Calculate the moles of NH4Cl:
moles of NH4Cl = 2.80 g / 53.49 g/mol ≈ 0.0524 mol
6. Calculate the dissolution enthalpy per mole:
ΔH_dissolution = q_total / moles of NH4Cl ≈ 0.885 kJ / 0.0524 mol ≈ 16.9 kJ/mol
Therefore, the dissolution enthalpy of NH4Cl(s) is approximately 16.9 kJ/mol. This positive value indicates that the dissolution process is endothermic, meaning it absorbs heat from the surroundings.
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