Please help me with this chem question

Question

How can a 0.50 M BaCl2 solution be diluted to yield one that contains 20.0 mg Ba² (the charge is 2 )/mL solution?

Answer 1

AW of Ba is 137.4, so the 0.50M soln contains 68.7 g/l (= 68.7 mg/ml) with respect to Ba++...the 2Cl- are of no importance.

To reduce this to 20.0 mg/ml, you need 68.7/20.0 = 3.43 vols of final soln...so add 2.43 vols of water to each vol of original soln...eg 2.43 l  of water to 1.00 l of soln.