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In the laboratory a general chemistry student finds that when 1.66 g CaBr2(s) is dissolved in 100.60 g water, the temperature of the solution increases from 23.95 to 26.02 °c. the heat capacity of the calorimeter was determined in a separate experiment to be 1.85 J/°C. Based on the student's observation, calculate the dissolution enthalpy of CaBr2(s) in kj/mol.
Accepted Answer
The dissolution enthalpy of CaBr2(s) can be calculated using the following steps:
1. Calculate the heat absorbed by the solution:
ΔT = 26.02 °C - 23.95 °C = 2.07 °C
q_solution = m_solution
C_solution
ΔT
m_solution = 100.60 g (water) + 1.66 g (CaBr2) = 102.26 g
C_solution = 4.184 J/g°C (specific heat of water)
q_solution = 102.26 g
4.184 J/g°C
2.07 °C = 883.9 J
2. Calculate the heat absorbed by the calorimeter:
q_calorimeter = C_calorimeter
ΔT
q_calorimeter = 1.85 J/°C
2.07 °C = 3.83 J
3. Calculate the total heat absorbed:
q_total = q_solution + q_calorimeter = 883.9 J + 3.83 J = 887.7 J
4. Calculate the moles of CaBr2:
moles_CaBr2 = mass_CaBr2 / molar_mass_CaBr2
moles_CaBr2 = 1.66 g / 199.89 g/mol = 0.00830 mol
5. Calculate the enthalpy change per mole of CaBr2:
ΔH_dissolution = q_total / moles_CaBr2
ΔH_dissolution = 887.7 J / 0.00830 mol = 106940 J/mol = 106.9 kJ/mol
Therefore, the dissolution enthalpy of CaBr2(s) is approximately 106.9 kJ/mol.
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