Answer:
8.78 g of magnesium acetate, Mg(CH₃COO)₂
Explanation:
From the question given above, the following data were obtained:
Molarity of Mg(CH₃COO)₂ = 0.247 M
Volume of solution = 250 mL
Mass of Mg(CH₃COO)₂ needed =?
Next, we shall convert 250 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
250 mL = 250 mL × 1 L / 1000 mL
250 mL = 0.25 L
Next, we shall determine the number of mole of Mg(CH₃COO)₂ in the solution. This can be obtained as follow:
Molarity of Mg(CH₃COO)₂ = 0.247 M
Volume of solution = 0.25 L
Mole of Mg(CH₃COO)₂ =?
Molarity = mole /Volume
0.247 = Mole of Mg(CH₃COO)₂ / 0.25
Cross multiply
Mole of Mg(CH₃COO)₂ = 0.247 × 0.25
Mole of Mg(CH₃COO)₂ = 0.0618 mole
Finally, we shall determine the mass of magnesium acetate, Mg(CH₃COO)₂, needed to prepare the solution. This can be obtained as follow:
Molar mass of Mg(CH₃COO)₂ = 24 + 2[12 + (3×1) + 12 + 16 + 16]
= 24 + 2[12 + 3 + 44]
= 24 +2[59]
= 24 + 118
= 142 g/mol
Mole of Mg(CH₃COO)₂ = 0.0618 mole
Mass of Mg(CH₃COO)₂ =?
Mole = mass / Molar mass
0.0618 = Mass of Mg(CH₃COO)₂ / 142
Cross multiply
Mass of Mg(CH₃COO)₂ = 0.0618 × 142
Mass of Mg(CH₃COO)₂ = 8.78 g
Thus, 8.78 g of magnesium acetate, Mg(CH₃COO)₂, is needed to prepare the solution.
