How many grams of aluminum nitrate are required to make 500.0 mL of a 0.0525 M aqueous solution?

 How many grams of aluminum nitrate are required to make 500.0 mL of a 0.0525 M aqueous solution?

(0.0525 mol/L) × (0.500 L) = 0.02625 mols

(0.02625 mol) × 213.0 g/mol) = 5.59 g

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