1 Answer

Ahmed Elashker · Answer 1 · 2021-05-10T23:33:41+0000

Answer:

The mass of FeSO₄.7H₂O needed
$\simeq$ 225.89 grams

Step-by-step 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 =
$(moles * molecular weight * 250\ mL)/(1000 \ mL)$

The mass of FeSO₄.7H₂O needed =
(3.25 * 278.02 * 250)/(1000)

The mass of FeSO₄.7H₂O needed = 225.89125 grams

The mass of FeSO₄.7H₂O needed
$\simeq$ 225.89 grams

Please log in or register to add a comment.