Question

6.01 g of mgso₄ is placed into 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 the water is 1.00 g/ml). you can assume that the specific heat of the solution is the same as that of water.

Answer 1

To calculate ∆H, the heat of solution for the dissolution of MgSO₄, we use q = mC∆T and convert mass to moles to calculate ∆H = q / n.

Step-by-step explanation:

To calculate ∆H, the heat of solution for the dissolution of MgSO₄, we can use the formula q = mC∆T, where q is the heat released or absorbed, m is the mass of the solution, C is the specific heat capacity, and ∆T is the change in temperature.

In this case, the mass of the solution is 100.0 g (since the density of water is 1.00 g/mL), the specific heat capacity is 4.184 J/g * °C (the same as water), and ∆T = 6.70 °C. We need to convert the mass of MgSO₄ to moles by dividing by its molar mass, which is 120.39 g/mol.

Then we can calculate ∆H by dividing the heat absorbed or released by the number of moles.

Using q = mC∆T, we can calculate q = (100.0 g) * (4.184 J/g * °C) * (6.70 °C) = 2799.968 J.

Converting grams of MgSO₄ to moles: (6.01 g) / (120.39 g/mol) = 0.0499 mol.

Therefore, ∆H = q / n = (2799.968 J) / (0.0499 mol) = 56.094 kJ/mol.