A solution is prepared by dissolving 0.1707 grams of (NH4)2Ce(NO3)6 in 1.46 L of water. Calculate the concentration of NO3- in this solution in mg/L (ppm).

Answer 1

molar mass (NH₄)₂Ce(NO₃)₆ = 548.2 g/mole

0.1707 g x 1 mol/548.2 g = 3.114x10^-4 moles (NH₄)₂Ce(NO₃)₆

3.114x10^-4 moles (NH₄)₂Ce(NO₃)₆ x 6 mol NO₃^-/mole (NH₄)₂Ce(NO₃)₆ = 1.868x10^-3 mol NO₃^-

molar mass NO3- = 62 g/mol

1.868x10^-3 mol NO₃^- x 62 g/mol = 0.1158 g NO₃^-

0.1158 g x 1000 mg/g = 115.8 mg NO₃^-

ppm = mg/L = 115.8 mg/1.46 L = 79.3 mg/L (ppm)