Calculate the concentration (in M) of NO3- in a solution prepared by dissolving 84.2 g Al(NO3)3 in water to prepare 456 mL of solution.

Answer 1

Concentration (in M) means moles solute / liter solution. We know there are 456 mls or solution = 0.456 liters of solution, so all we need to find is the moles of NO₃^- that are present in 84.2 g of Al(NO₃)₃.

moles Al(NO₃)₃ = 84.2 g x 1 mol / 213 g = 0.395 moles

moles of NO₃^- = 0.395 moles Al(NO₃)₃ x 3 moles NO₃^- / mole Al(NO₃)₃ = 1.185 moles NO₃^-

Molarity (M) = 1.185 moles / 0.456 L = 2.60 M NO₃^-