I don't know how to answer this question, plz help!

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

[Ba2+] = 0.50 mol/L = 0.50 x 137.33 g/mol/1L= 68.665 g/L =

= 0.0687 g/mL =>  68.7 mg/mL

68.7 / 20.0 = 3.43

For each mL of 0.50 M you must add 3.43 - 1= 2.43 mL