Molarity told you how many moles of a compound are in each liter. Electrolytes and Ions in Solution told you how many ions each formula unit releases. Put them together and you get the concentration of each ion, the number that matters for precipitation, colligative properties and equilibrium.
Key idea
$$[\text{ion}] = (\text{molarity of the compound}) \times (\text{ions of that kind per formula unit})$$ So $0.20\ \text{M}\ \mathrm{CaCl_2}$ is $0.20\ \text{M}$ in $\mathrm{Ca^{2+}}$ and $0.40\ \text{M}$ in $\mathrm{Cl^-}$. When solutions are mixed, add the moles of each ion and divide by the total volume.
Why it works
Every formula unit of $\mathrm{CaCl_2}$ that dissolves gives one calcium ion and two chloride ions. So a liter that received 0.20 mol of $\mathrm{CaCl_2}$ holds 0.20 mol of $\mathrm{Ca^{2+}}$ and 0.40 mol of $\mathrm{Cl^-}$. Concentrations can't be added directly when you mix, because each solution gets diluted by the other's volume; moles can be added, since ions don't disappear when solutions are poured together (unless they react).
Worked example: Two ions from one salt
Find $[\mathrm{Al^{3+}}]$ and $[\mathrm{SO_4^{2-}}]$ in $0.150\ \text{M}\ \mathrm{Al_2(SO_4)_3}$.
- One formula unit gives 2 $\mathrm{Al^{3+}}$ and 3 $\mathrm{SO_4^{2-}}$.
- $[\mathrm{Al^{3+}}] = 2 \times 0.150 = 0.300\ \text{M}$; $[\mathrm{SO_4^{2-}}] = 3 \times 0.150 = 0.450\ \text{M}$.
Answer: $0.300\ \text{M}$ and $0.450\ \text{M}$.
Worked example: From grams
$2.22\ \text{g}$ of $\mathrm{CaCl_2}$ ($110.98\ \text{g/mol}$) is dissolved to make $250.0\ \text{mL}$ of solution. Find $[\mathrm{Cl^-}]$.
- Moles: $2.22/110.98 = 0.02000\ \text{mol}$; molarity $0.02000/0.2500 = 0.0800\ \text{M}$.
- $[\mathrm{Cl^-}] = 2 \times 0.0800 = 0.160\ \text{M}$.
Answer: $0.160\ \text{M}$
Worked example: Mixing two solutions
$100.0\ \text{mL}$ of $0.200\ \text{M}$ NaCl is mixed with $50.0\ \text{mL}$ of $0.100\ \text{M}\ \mathrm{CaCl_2}$. Find $[\mathrm{Cl^-}]$ in the mixture.
- Moles of $\mathrm{Cl^-}$: from NaCl, $0.1000 \times 0.200 = 0.0200$; from $\mathrm{CaCl_2}$, $0.0500 \times 0.100 \times 2 = 0.0100$.
- Total: $0.0300\ \text{mol}$ in $0.1500\ \text{L}$.
- $[\mathrm{Cl^-}] = 0.0300/0.1500 = 0.200\ \text{M}$.
Answer: $0.200\ \text{M}$
Ion concentrations are what the solution "knows". A mixture of NaCl and KCl, or of $\mathrm{CaCl_2}$ and $\mathrm{Ca(NO_3)_2}$, is described completely by the concentration of each ion; which salt each ion came from no longer matters. That is why net ionic equations, precipitation and freezing points are all worked out from ion concentrations.
Common mistake
Giving the chloride concentration as the salt's molarity, ignoring the subscript: "$0.20\ \text{M}\ \mathrm{CaCl_2}$ has $0.20\ \text{M}\ \mathrm{Cl^-}$." There are two chloride ions per formula unit, so it is $0.40\ \text{M}$. Check: for a neutral salt the positive and negative charges in solution must balance. Here $2(0.20) = 0.40$ of positive charge per liter matches $1(0.40)$ of negative.