To what volume should you dilute 135 mL of an 810 M CuCl2 so

To what volume should you dilute 135 mL of an 8.10 M CuCl2 solution so that 51.5 mL of the diluted solution contains 4.57 g CuCl2?

Solution

calculate final concentration

Molar mass of CuCl2,
MM = 1*MM(Cu) + 2*MM(Cl)
= 1*63.55 + 2*35.45
= 134.45 g/mol


mass(CuCl2)= 4.57 g

use:
number of mol of CuCl2,
n = mass of CuCl2/molar mass of CuCl2
=(4.57 g)/(134.45 g/mol)
= 3.399*10^-2 mol
volume , V = 51.5 mL
= 5.15*10^-2 L


use:
Molarity,
M = number of mol / volume in L
= 3.399*10^-2/5.15*10^-2
= 0.660 M

use dilution formula
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution

Given:
M1 = 8.1 M
M2 = 0.66 M
V1 = 135.0 mL

use:
M1*V1 = M2*V2
V2 = (M1 * V1) / M2
V2 = (8.1*135)/0.66
V2 = 1657 mL
V2 = 1.66*10^3 mL
Answer: 1.66*10^3 mL