Question

A chemist makes a solution of Mg(NO3)2 by dissolving 20.9 g Mg(NO3)2 in water to make 100.0 mL of solution. What is the concentration of NO3− ions in the solution? Assume that Mg(NO3)2 is the only solute in the solution. The molar mass of Mg(NO3)2 is 148.33 g/mol.

Answer 1

Mass of Mg(NO₃)₂ used for preparing solution = 20.9 g. The molar mass of Mg(NO₃)₂ is 148.33 g/mol.

Hence, the number of moles of Mg(NO₃)₂ used for preparing solution = mass/molar mass = 20.9 g/ 148.33 g/mol = 0.141 mol.

1 mole of Mg(NO₃)₂ furnishes 2 moles of NO₃^- ion in solution.

Hence the number of moles of NO₃^- furnished in the solution by 0.141 mol of Mg(NO₃)₂ = 2 X 0.141 = 0.282 mol.

The volume of the solution = 100.0 mL = 0.100 L.

Hence, the concentration of NO₃^- ion = 0.282 mol / 0.100 L = 2.82 M.