# A student dissolves 4.00 g of anhydrous copper(II) sulfate in 50.0 cm³ of water and measures a temperature rise of 14.2 °C. Explain how the student would determine the enthalpy change of dissolution per mole of copper(II) sulfate from this experiment, and identify one significant source of error that would make the calculated value less negative than the true value.

> IB DP Chemistry Higher Level (2023 syllabus) — R1.1 Measuring enthalpy changes · Explain · 4 marks

> Specific heat capacity of the solution = 4.18 J g⁻¹ K⁻¹. Molar mass of CuSO₄ = 159.6 g mol⁻¹. Assume the density of the solution is 1.00 g cm⁻³.

## Mark scheme (4 marks)

1. The heat released by the reaction (q) is calculated using q = mcΔT, where m is the mass of the solution (approximately 50.0 g), c is the specific heat capacity, and ΔT is the temperature change.
2. The enthalpy change per mole is obtained by dividing q by the number of moles of CuSO₄ dissolved (n = 4.00/159.6 ≈ 0.02506 mol), giving ΔH in J mol⁻¹, then converting to kJ mol⁻¹.
3. Since the temperature rises, the dissolution is exothermic, so ΔH is assigned a negative sign (ΔH = −q/n).
4. A significant source of error is heat loss to the surroundings (e.g. to the air or the calorimeter/container), which means less heat is recorded by the solution, so the measured temperature rise is smaller than it should be, making the calculated ΔH less negative than the true value.

## Key terms

- [enthalpy change of dissolution](https://www.gradenine.co.uk/glossary/enthalpy-change-of-dissolution)

## Related

- [Revision notes for IB DP Chemistry Higher Level (2023 syllabus)](https://www.gradenine.co.uk/learn)
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Source: [GradeNine](https://www.gradenine.co.uk/q/a-student-dissolves-4-00-g-of-fe3bd6e9) · Published by Druglandscape Ltd.