I found a similar problem to this as shown in the attached picture to helpme answer the problem. We use the M₁V₁ = M₂V₂ formula, wherein M₁ is the concentration of the original CuSO₄ solution, V₁ is the volume of CuSO₄ in the cuvette, M₂ is the concentration of Cu²⁺ in the cuvette, and V₂ is the total volume in the cuvette.
For cuvette 1:
M₁V₁ = M₂V₂
(0.06)(4 mL) = (M₂)(6+4 mL)
M₂ = 0.024 M Cu²⁺
For cuvette 2:
M₁V₁ = M₂V₂
(0.06)(7 mL) = (M₂)(7+3 mL)
M₂ = 0.042 M Cu²⁺
