Answer:
4.1 × 10⁻⁵ M
Explanation:
Let's consider the following balanced equation.
FeCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Fe(NO₃)₃(aq)
We can establish the following relations:
When 3.5 mg of AgCl are collected, the moles of of FeCl₃ that reacted are:
[tex]3.5 \times 10^{-3} gAgCl.\frac{1molAgCl}{143.32gAgCl} .\frac{1molFeCl_{3}}{3molAgCl} =8.1 \times 10^{-6} molFeCl_{3}[/tex]
The sample has a volume of 200 mL (0.200 L). The molar concentration of FeCl₃ is:
[tex]\frac{8.1 \times 10^{-6} molFeCl_{3}}{0.200L} =4.1 \times 10^{-5}M[/tex]