Question

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

Answer 1

`"Concentration"` `=` `("Moles of Solute")/("Volume of Solution")`

Explanation:

Quite clearly, if I have 2 solutions, EACH WITH EQUAL CONCENTRATIONS of REAGENT, say `1` `mol` `L^-1`, the solution of calcium chloride will have greater ionic strength than that of sodium chloride, in that `[NaCl]` is `1` `mol` `L^-1` with respect to `[Na^+]` and `[Cl^-]`, whereas `[CaCl_2]` is `1` `mol` `L^-1` with respect to `[Na^+]`, BUT `2` `mol` `L^-1` with respect to chloride `[Cl^-]`. 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 `CaCl_2` is `1` `mol` `L^-1`; the `[Cl^-]` concentration is `2` `mol` `L^-1`.