How does chloride ion concentration compare given 1*mol*L^-1 solutions EACH of NaCl, and CaCl_2?

$\text{Concentration}$ $=$ $\left(\text{Moles of Solute")/("Volume of Solution}\right)$
Quite clearly, if I have 2 solutions, EACH WITH EQUAL CONCENTRATIONS of REAGENT, say $1$ $m o l$ ${L}^{-} 1$, the solution of calcium chloride will have greater ionic strength than that of sodium chloride, in that $\left[N a C l\right]$ is $1$ $m o l$ ${L}^{-} 1$ with respect to $\left[N {a}^{+}\right]$ and $\left[C {l}^{-}\right]$, whereas $\left[C a C {l}_{2}\right]$ is $1$ $m o l$ ${L}^{-} 1$ with respect to $\left[N {a}^{+}\right]$, BUT $2$ $m o l$ ${L}^{-} 1$ with respect to chloride $\left[C {l}^{-}\right]$. We deal with the number of solute particles.
I don't know how this relates to the tonicity of the spud. But if I report that the concentration of $C a C {l}_{2}$ is $1$ $m o l$ ${L}^{-} 1$; the $\left[C {l}^{-}\right]$ concentration is $2$ $m o l$ ${L}^{-} 1$.