Question

You need to make an aqueous solution of 0.162 M barium chloride for an experiment in lab, using a 500 mL volumetric flask. How much solid barium chlorideshould you add

Answer 1

Answer

Molarity= moles of solute /volume of solution L

Molarity= 0.162 M and volume of solution = 500 ml= 0.5 L

moles of solute= Molarity*volume of solution in L = 0.162*0.5= 0.081 moles of barium chloride

mass of barium chloride = moles*molar mass= 0.081*208.23= 16.9 grams