54.20 g of Al(NO3)3 in 1500 mL of solution

Answer 1

What's the question? Are you asking for the molarity of the solution?

molar mass Al(NO3)3 = 212.99 g/mol

54.20 g x 1 mol / 212.99 g = 0.2545 moles

molarity = moles / liter

molarity (M) = 0.2545 moles / 1.500 L = 0.1696 M = 0.17 M (2 sig. figs. based on 1500 ml having 2 s.f.)