Question

How many grams of magnesium acetate are required to prepare a 50.0 ml of 0.100 M acetate ion solution?

Answer 1

Molar mass of Mg(CH3COO)2 = 142.4 g/mol

Molarity M = moles of solute x 1000 / volume of solution

or, moles of solute Mg(CH3COO)2 = Molarity x volume of solution / 1000

moles of solute Mg(CH3COO)2 = 0.100 x 50.0 / 1000

moles of solute Mg(CH3COO)2 = 0.005 mole

Now, mass related to 0.005 mole of Mg(CH3COO)2 = 0.005 x 142.4

= 0.712g

which will give 0.200M of acetate ion as one mole of Mg(CH3COO)2 gives two moles of CH3COO- ions

Hence, amount of Mg(CH3COO)2 to get 0.100M of acetate ion = 0.712 / 2

= 0.356g of Mg(CH3COO)2