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

Mg(NO3)2 ➡️ Mg2+ + 2 NO3-

（32.0g Mg(NO3)2) / (148.3g Mg(NO3)2/mol)* (2 mol NO3- / 1 mol Mg(NO3)2) / (0.425 L) = 1.02 mol/ L NO3-