Question

You need to make an aqueous solution of 0.125 M barium chloride for an experiment in lab, using a 250 mL volumetric flask. How much solid barium chloride should you add?

Answer 1

6.51g

Step-by-step explanation:

0.125 M = 0.125 mol/1000ml

0.125/4 = 0.03125 mol

BaCl2:

Mr = 137.3 + 2(35.5) = 208.3

mass = 208.3(0.03125) = 6.509375