Ion Concentrations

In $0.2\ \mathrm{M}$ CaCl₂ no CaCl₂ is left whole: each one became one Ca²⁺ and two Cl⁻.

CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq)

So [Ca²⁺] = $0.2\ \mathrm{M}$ and [Cl⁻] = $0.4\ \mathrm{M}$. Square brackets mean "the concentration of".

[Ca²⁺] = 0.2 M, [Cl⁻] = 0.4 M

Multiply the concentration of the solid by the number of that ion in the dissolution equation. The volume does not change the concentrations.