Question

How many grams of CaCl2 are present in 28.0 mL of a 0.520 M solution? Express your answer using three significant figures

Answer 1

Molarity = 0.520 M

Volume = 28 ml = 28/1000 = 0.028 lts

No. of moles of CaCl₂ = Molarity x Volume in lts = 0.52 M x 0.028 lts = 0.0146 moles

Molar mass of CaCl₂ = 111 g/mol

Mass of CaCl₂ = Moles x molar mass = 0.0146 mol x 111 g/mol = 1.62 gms (three significant figures)