Answer:
You need 0.702g of ferrous ammonium sulfate hexahydrate to prepare the desired solution.
Explanation:
First, you want to prepare 250.0mL of a 400.0ppm of Fe²⁺ (That is mg/L, w/v):
0.250L * (400mg / L) = 100mg Fe²⁺ = 0.1g Fe²⁺
Molar mass of Fe is 55.845g. moles are:
0.1g Fe²⁺ * (1mol / 55.845g) =
<h3>1.79x10⁻³ moles Fe²⁺</h3>
The Fe²⁺ comes from ferrous ammonium sulfate hexahydrate, as 1 mole of this salt contains 1 mole of Fe²⁺, moles of the salt you need are:
1.79x10⁻³ moles Fe(NH₄)₂(SO₄)₂.6H₂O.
To convert these moles to grams you must use molar weight, MW, thus:
1.79x10⁻³ moles Fe(NH₄)₂(SO₄)₂.6H₂O * (392.14g / 1mol) =
<h3>You need 0.702g of ferrous ammonium sulfate hexahydrate to prepare the desired solution.</h3>
