What mass of MgSO4 · 7 H2O is required to

prepare 300 mL of a 0.511 M MgSO4 solution?

See your duplicate post above.

To determine the mass of MgSO4 · 7 H2O required to prepare the solution, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

Given:
Molarity (M) = 0.511 M
Volume of solution = 300 mL = 0.300 L

First, we need to calculate the moles of MgSO4 in the solution. Rearranging the formula, we have:

moles of solute = Molarity × volume of solution

moles of MgSO4 = 0.511 M × 0.300 L = 0.1533 moles

Next, we need to find the molar mass of MgSO4 · 7 H2O, which consists of MgSO4 and 7 molecules of water (H2O). The molar mass for each component is:

Molar mass of MgSO4 = 24.31 g/mol + 32.06 g/mol + (4 × 16.00 g/mol) = 120.37 g/mol
Molar mass of H2O = 2 × 1.01 g/mol + 16.00 g/mol = 18.02 g/mol

Now, we can calculate the total molar mass of MgSO4 · 7 H2O:

Molar mass of MgSO4 · 7 H2O = Molar mass of MgSO4 + (7 × Molar mass of H2O)
= 120.37 g/mol + (7 × 18.02 g/mol)
= 246.29 g/mol

Finally, to find the mass of MgSO4 · 7 H2O, we multiply the moles of MgSO4 by the molar mass of MgSO4 · 7 H2O:

Mass of MgSO4 · 7 H2O = Moles of MgSO4 × Molar mass of MgSO4 · 7 H2O
= 0.1533 moles × 246.29 g/mol
= 37.74 grams

Therefore, approximately 37.74 grams of MgSO4 · 7 H2O is required to prepare 300 mL of a 0.511 M MgSO4 solution.