to what volume should you dilute 125mL of an 8.00M CuCl2 solution so that 50.0mL of the diluted solution contains 5.9g CuCl2

Question

I have a hard time determing the
M1V1=M2V2 is

Answer 1

molar mass CuCl2 = 134.45
8.00M x 0.125L = 1.00 mole CuCl2 in 125 mL. This is 134.45g/125 mL = 1.0756 g/mL in hand.
We want 5.9g/50 mL or 0.118 g/mL.
We want to dilute by a factor of
1.0756/0.118 = 9.11625 (I know that's too many significant figures).
Then 9.11525 x 125 mL = total volume of 1139.407. Diluted, that soln now is
1.0756g/mL x (125/1139.407) = 0.118 g/mL.
If we take 50 mL of that soln we will have 0.118g/mL x 50 mL = 5.90grams CuCl2.
So the answer is to dilute the 125 mL of 8.00M CuCl2 to 1139 mL. Check my work; it's late here.