Molarity Practice

Question

7. Calculate the mass of glucose (C₆H₁₂O₆) in a 155 mL sample of a 1.18 M glucose solution.        

Answer 1

molar mass glucose = 180.2 g/mol

155 ml x 1 L/1000 ml = 0.155 L

1.18 M = 1.18 mols glucose / L

moles glucose present = 1.18 mols / L x 0.155 L = 0.1829 mols

mass of glucose = 0.1829 mols x 180.2 g / mol = 32.96 g glucose = 33.0 g glucose (3 sig. figs.)