Question

What is the concentration of nitrate ion in a 425 ml solution containing 32.0 g of Mg(NO3)2 (M = 148.3)

Answer 1

Supposing complete ionization:
Mg(NO3)2 → Mg{2+} + 2 NO3{-}

(32.0 g Mg(NO3)2) / (148 g Mg(NO3)2/mol) x (2 mol NO3{-} / 1 mol Mg(NO3)2) / (0.425 L) = 1.02 mol/L NO3{-}