Molarity Practice

Question

9. What volume of a 0.31 M Mg(NO₃)₂ solution contains 88 g of Mg(NO₃)₂?

Answer 1

molar mass Mg(NO₃)₂ = 148.3 g/mol

how many moles are in 88 g? 88 g x 1 mol / 148.3 g = 0.593 mols Mg(NO₃)₂

0.31 M = 0.31 mols Mg(NO₃)₂ / L

(x L) (0.31 mols / L) = 0.593 mols

x = 1.91 L contains 0.593 mols which is 88 g of Mg(NO₃)₂