Answer:
[tex]\large \boxed{1.44 \times 10^{-5} \text{ mol/L}}[/tex]
Explanation:
CaI₂(s) ⟶ Ca²⁺ + 2I⁻
1. Original concentration of I⁻
The chemist diluted 20 mL of the original solution to 100 mL
We can use the dilution formula to calculate the original concentration of I⁻.
[tex]\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{20 mL}\times c_{1} & = & \text{100 mL} \times 5.74 \times 10^{-6} \text{ mol/L}\\20c_{1}& = & 5.74 \times 10^{-4} \text{ mol/L}\\c_{1}& = & 2.87 \times 10^{-5} \text{ mol/L}\\\end{array}[/tex]
2. Original concentration of Ca²⁺
The molar ratio is 1 mol Ca²⁺:2 mol I⁻.
[tex][\text{Ca}^{2+}] =\dfrac{2.87 \times 10^{-5} \text{ mol I}^{-}}{\text{1 L}} \times \dfrac{\text{1 mol Ca}^{2+} }{\text{2 mol I}^{-}} = \mathbf{1.44 \times 10^{-5}} \textbf{ mol/L}\\\\\text{[Ca$^{2+}$] in the original solution was $\large \boxed{\mathbf{1.44 \times 10^{-5}} \textbf{ mol/L}}$}[/tex]
