To what volume should you dilute 131 mL of an 8.20 M CuCl2 solution so that 49.0 mL of the diluted solution contains 4.53g CuCl2 ?

4 answers

Molar mass of CuCl2 is 134.5 g mol^-1

so solution needs to contain

4.53 g / 134.5 g mol^-1
=0.03368 mole

concentration is

0.049 L/0.03368 mole
=1.455 molar

thus we need to dilute the 8.20 M solution by a factor
8.20 M/1.455 M= 5.636

so the new total volume is

131 ml x 5.636
=738.4 ml

check on the concentration

old solution contains
131 x 8.20 mmoles = 1074 mmoles

new solution contains
1.455 x 738.4 mmole = 1074 mmole
Thank you for all the detail, but i'm still confused......so 738.4mL is the answer?
Thank you :)
this answer is incorrect. The following should be switched:

0.049 L/0.03368 mole
=1.455 molar

should be
0.03368 mol/0.049 L
=0.6873 molar

the rest of the steps are correct but you would need to substitute the above number in for the remainder of the problem