Answer:
14,8mg of Copper
Explanation:
Lambert-Beer law says:
A = abC
Where A is absorbance (0,874-0,056 = 0,818), a is molar absorption coefficient (7,90x10³M⁻¹cm⁻¹), b is path length (1,0 cm) and C is molar concentration.
Solving for C:
C = 1,04x10⁻⁴M of (neocuproine)₂Cu+ = 1,04x10⁻⁴M Cu
This concentration is the concentration of Copper measured using the absorbance.
The concentration of Copper in solution B is:
1,04x10⁻⁴M Copper×[tex]\frac{20,00mL+10,00mL+15,00mL}{15,00mL}[/tex] = 3,11x10⁻⁴M Copper
Copper in solution A is:
3,11x10⁻⁴M Copper×[tex]\frac{10,00mL+10,00mL+10,00mL}{10,00mL}[/tex] = 9,32x10⁻⁴M Copper
As the total volume of solution A is 250,0mL, moles of Copper in A are:
9,32x10⁻⁴M Cu × 0,2500L = 2,33x10⁻⁴ moles of Copper
In grams:
2,33x10⁻⁴ moles of Cu×[tex]\frac{63,546g}{1mol}[/tex] = 0,0148 g of Copper ≡ 14,8 mg of Copper in an unknown
I hope it helps!