Calculation of grams of magnesium bromide needed:

To determine the number of grams of magnesium bromide needed to make a 4.0M solution, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (L)

Step 1: Convert molarity to moles:
Given that the molarity is 4.0M and the volume of the solution is 1.0L, we can use the formula to find the moles of magnesium bromide:

Moles of solute = Molarity × Volume of solution

Moles of solute = 4.0M × 1.0L = 4.0 moles

Step 2: Convert moles to grams:
To convert moles of magnesium bromide to grams, we need the molar mass of magnesium bromide (MgBr2). The molar mass of magnesium (Mg) is 24.31 g/mol, and the molar mass of bromine (Br) is 79.90 g/mol.

The molar mass of magnesium bromide (MgBr2) is:
(24.31 g/mol × 1) + (79.90 g/mol × 2) = 183.11 g/mol

Now we can use the formula:

Mass (grams) = Moles × Molar mass

Mass (grams) = 4.0 moles × 183.11 g/mol = 732.44 grams

Therefore, 732.44 grams of magnesium bromide are needed to make a 1.0L solution with a molarity of 4.0M.

Summary:
To make a 1.0L solution of 4.0M magnesium bromide, we would need 732.44 grams of magnesium bromide. This calculation was done by first converting the molarity to moles using the formula Molarity = moles/volume. Then, the moles were converted to grams using the molar mass of magnesium bromide. The molar mass of magnesium bromide is the sum of the molar masses of magnesium and bromine.