how many grams of iron(ii) sulfate heptahydrate is needed to prepare 250.0 ml of a 3.25 molar aqueous solution

The mass of FeSO₄.7H₂O  needed  225.89 grams

Explanation:

From the given question

The molecular weight of iron(ii) sulfate heptahydrate FeSO₄.7H₂O = 278.02 g/mol

Molarity = 3.25

Volume = 250.0 mL

The mass of FeSO₄.7H₂O  needed =

The mass of FeSO₄.7H₂O  needed  =

The mass of FeSO₄.7H₂O  needed  = 225.89125 grams

The mass of FeSO₄.7H₂O  needed  225.89 grams