Answer:
6,04x10⁻³M
Explanation:
For the reaction:
Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
The precipitate of Cu(s) weights 96,0 mg. In moles:
Moles of Cu(s):
0,096g×(1mol/63,546g) = 1,51x10⁻³ moles of Cu(s). If you see the balanced equation 1 mole of CuSO₄ produce 1 mole of Cu(s). That means moles of CuSO₄ are the same of Cu(s), 1,51x10⁻³ moles of CuSO₄
As volume of the solution is 250 mL, 0,250L, the molar concentration of the original solution is:
1,51x10⁻³ moles of CuSO₄ / 0,250L = 6,04x10⁻³M
I hope it helps!