Question

8) In the laboratory you dissolve 21.8 g of magnesium iodide in a volumetric flask and add water to a total volume of 125 mL. What is the molarity of the solution? M. What is the concentration of the magnesium cation? M. What is the concentration of the iodide anion? M.

Answer 1

The molecular formula of magnesium iodide = MgI₂

The mass of MgI₂ dissolved in water = 21.8 g

The molar mass of MgI₂ = 278 g/mol

Now, the no. of moles of MgI₂ = 21.8 g / 278 g mol^-1 = 0.078 mol

The volume of the solution = 125 mL = 125/1000 = 0.125 L (since 1 L = 1000 mL)

Formula: Molarity of MgI₂ = no. of moles of MgI₂ / volume of solution (L)

i.e. The molarity of solution of MgI₂ = 0.627 M

Since MgI2 is a strong electrolyte, it dissociates completely in solution, as shown below.

MgI₂(aq) Mg²⁺(aq) + 2I^-(aq)

i.e. [MgI₂] = [Mg²⁺] = 2[I^-]

i.e. [Mg²⁺] = 0.627 M and [I^-] = 0.627/2 = 0.314 M