A solution is prepared by dissolving 0.0627 grams of (NH4)2Ce(NO3)6 in 1.72 L of water. Calculate the concentration of NO3- in this solution in mg/L (ppm).

Answer 1

molar mass (NH₄)₂Ce(NO₃)₆ = 584.2 g/mole

moles of (NH₄)₂Ce(NO₃)₆ = 0.0627 g x 1 mol/584.2 g = 1.07x10^-4 moles

moles NO₃^- = 1.07x10^-4 moles (NH₄)₂Ce(NO₃)₆ x 6 mol NO₃^- / mole (NH₄)₂Ce(NO₃)₆ = 6.42x10^-4 mol

6.42x10-4 mol NO3- / 1.72 L = 3.73x10-4 mol / L

molar mass NO₃^- = 62 g/mol

3.73x10^-4 mol / L x 62 g/mol = 2.31x10^-2 g/L NO₃^- = 0.0231 g/L = 23.1 mg/L = 23.1 ppm NO₃^-