You need to make an aqueous solution of 0.131 M iron(II) acetate for an experiment in lab, using a 125 mL volumetric flask. How much solid iron(II) acetate should you add

2.85 g of iron(II) acetate, Fe(C₂H₃O₂)₂

Explanation:

The following data were obtained from the question:

Molarity of Fe(C₂H₃O₂)₂ = 0.131 M

Volume of solution = 125 mL

Mass of Fe(C₂H₃O₂)₂ =?

Next, we shall convert 125 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

125 mL = 125 mL × 1 L / 1000 mL

125 mL = 0.125 L

Thus, 125 mL is equivalent to 0.125 L.

Next, we shall determine the number of mole of Fe(C₂H₃O₂)₂ in the solution. This can be obtained as follow:

Molarity of Fe(C₂H₃O₂)₂ = 0.131 M

Volume of solution = 0.125 L

Mole of Fe(C₂H₃O₂)₂ =?

Molarity = mole /Volume

0.131 = Mole of Fe(C₂H₃O₂)₂ / 0.125

Cross multiply

Mole of Fe(C₂H₃O₂)₂ = 0.131 × 0.125

Mole of Fe(C₂H₃O₂)₂ = 0.0164 mole

Finally, we shall determine the mass of iron(II) acetate, Fe(C₂H₃O₂)₂, needed to prepare the solution. This can be obtained as follow:

Mole of Fe(C₂H₃O₂)₂ = 0.0164 mole

Molar mass of Fe(C₂H₃O₂)₂ = 56 + 2[(2×12) + (3×1) + (2×16)]

= 56 + 2[24 + 3 + 32]

= 56 + 2[59]

= 56 + 118

Molar mass of Fe(C₂H₃O₂)₂ = 174 g/mol

Mass of Fe(C₂H₃O₂)₂ =?

Mole = mass /Molar mass

0.0164 = Mass of Fe(C₂H₃O₂)₂ / 174

Cross multiply

Mass of Fe(C₂H₃O₂)₂ = 0.0164 × 174

Mass of Fe(C₂H₃O₂)₂ = 2.85 g

Thus, to prepare 0.131 M iron(II) acetate, Fe(C₂H₃O₂)₂, add 2.85 g of Fe(C₂H₃O₂)₂ to 125 mL volumetric flask and fill with water to the mark.