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

1 Answer
Mar 5, 2016

Answer:

#"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#.