You need to make an aqueous solution of 0.247 M magnesium acetate for an experiment in lab, using a 250 mL volumetric flask. How much solid magnesium acetate should you add

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.

the answer is barium and copper bromide