Molarity Practice

Question

8. A laboratory procedure calls for making 580.0 mL of a 1.2 M KNO₃ solution.

How much KNO₃ in grams is needed?        

Answer 1

580.0 ml x 1 L/1000 ml = 0.5800 L

1.2 M = 1.2 mols KNO₃ / L

1.2 mols KNO₃ / L x 0.5800 L = 0.696 mols KNO₃ required

Molar mass KNO₃ = 101.1 g/mol

grams needed = 0.696 mols x 101.1 g/mol = 70.37 g = 70 g KNO₃ (2 sig. figs.)