What mass of MgSO4 7 H2O is required to prepare 300 mL of a 0.529 M MgSO4 solution?

To prepare a 0.529 M MgSO4 solution, you first need to calculate the moles of MgSO4 required for 300 mL (0.300 L) of solution:

Moles of MgSO4 = Molarity × Volume = 0.529 mol/L × 0.300 L = 0.1587 mol.

Next, since MgSO4·7H2O is the hydrated form, its molar mass is approximately 246.47 g/mol. Therefore, the mass needed is:

Mass = Moles × Molar Mass = 0.1587 mol × 246.47 g/mol ≈ 39.06 g of MgSO4·7H2O.