A solution was prepared by dissolving 7.45 g of KCl in 200 g of in a calorimeter with a water equiva — Thermodynamics and Thermochemistry Chemistry Question
Question
A solution was prepared by dissolving 7.45 g of KCl in 200 g of $\text{H}_2\text{O}$ in a calorimeter with a water equivalent of 25 g. The temperature of the water had reduced from $28^\circ\text{C}$ to $25^\circ\text{C}$. $\Delta H$ (in kJ/mol) for dissolving KCl in water is (Specific heat capacity of water is 4.2 J/K-g.)
💡 Solution & Explanation
Molar mass of KCl is 74.5 g/mol, so 7.45 g = 0.1 mol. Total mass heated equivalent = $200\text{ g (water)} + 25\text{ g (calorimeter)} = 225\text{ g}$. Heat absorbed from the system $q = mc\Delta T = 225 \times 4.2 \times (28 - 25) = 2835\text{ J}$. The enthalpy change per mole $\Delta H = +2835\text{ J} / 0.1\text{ mol} = 28350\text{ J/mol} = +28.35\text{ kJ/mol}$.