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.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.