Question

4.26 g of MgSO₄ is added to 100.0 ml of water. The water's temperature increases by 6.70 °C. Calculate ∆H, in kJ/mol, for the dissolution of MgSO₄. (The specific heat of water is 4.184 J/g･°C, and the density of water is 1.00 g/ml). Assume that the specific heat of the solution is the same as that of water.

Answer 1

The calculation involves finding the heat absorbed by the solution and then determining the enthalpy change per mole for the dissolution of MgSO₄ in water using the equations of calorimetry and stoichiometry.

Step-by-step explanation:

The question involves a thermal calorimetry experiment to determine the enthalpy change (ΔH) for the dissolution of MgSO₄ in water.

First, we need to calculate the amount of heat absorbed by the water when MgSO₄ is dissolved, using the formula q = m × C × ΔT, where m is the mass of the solution, C is the specific heat capacity (which is the same as water's), and ΔT is the change in temperature.

Since the density of water is 1.00 g/mL, the mass of 100.0 mL water is 100.0 g, so the total mass of the solution is 100.0 g + 4.26 g of MgSO₄. Now, we use the specific heat capacity (4.184 J/g°C) and the temperature increase (6.70°C) to find the heat absorbed: q = (100.0 g + 4.26 g) × 4.184 J/g°C × 6.70°C.

After calculating q, which gives us the heat in joules, we convert it to kilojoules and then divide by the number of moles of MgSO₄ to find ΔH per mole. The number of moles of MgSO₄ is calculated by dividing its mass by its molar mass.