Answer:
The mass of FeSO₄.7H₂O needed [tex]\simeq[/tex] 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 = [tex]\dfrac{moles \times molecular weight \times 250\ mL}{1000 \ mL}[/tex]
The mass of FeSO₄.7H₂O needed = [tex]\dfrac{3.25 \times 278.02 \times 250}{1000}[/tex]
The mass of FeSO₄.7H₂O needed = 225.89125 grams
The mass of FeSO₄.7H₂O needed [tex]\simeq[/tex] 225.89 grams
