Answer:
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.
