Calculate how many grams of FeSO4 . 7 H2O (MW = 278.02 g/mol) will be needed to prepare 1 mL of a 0.2 M FeSO4 solution?

Am I right when I say we'd need

.0304g?

To calculate the number of grams of FeSO4 . 7H2O needed to prepare 1 mL of a 0.2 M FeSO4 solution, we need to follow these steps:

Step 1: Find the molar mass of FeSO4 . 7H2O.
Given: Molar mass of FeSO4 . 7H2O (MW) = 278.02 g/mol.

Step 2: Calculate the number of moles required.
Given: Volume (V) = 1 mL = 0.001 L.
Molarity (M) = 0.2 M.

Using the formula: Moles (n) = Molarity (M) x Volume (V)
n = 0.2 mol/L x 0.001 L = 0.0002 mol

Step 3: Convert moles to grams.
Given: Moles (n) = 0.0002 mol.

Using the formula: Mass (m) = Moles (n) x Molecular Weight (MW)
m = 0.0002 mol x 278.02 g/mol
m = 0.0556 g

Therefore, you would need approximately 0.0556 grams of FeSO4 . 7H2O to prepare 1 mL of a 0.2 M FeSO4 solution.