How does chloride ion concentration compare given $1 \cdot m o l \cdot L^{- 1}$ $1molL^-1$ solutions EACH of $N a C l$ $NaCl$ , and $C a C l_{2}$ $CaCl_2$ ?

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How does chloride ion concentration compare given $1 \cdot m o l \cdot L^{- 1}$ $1molL^-1$ solutions EACH of $N a C l$ $NaCl$ , and $C a C l_{2}$ $CaCl_2$ ?

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Answer 1 · 2016-03-05T08:29:48

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of Solute}{Volume of Solution}$ $("Moles of Solute")/("Volume of Solution")$

Explanation:

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

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How does chloride ion concentration compare given $1 \cdot m o l \cdot L^{- 1}$ $1molL^-1$ solutions EACH of $N a C l$ $NaCl$ , and $C a C l_{2}$ $CaCl_2$ ?

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Explanation:

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How does chloride ion concentration compare given $1 \cdot m o l \cdot L^{- 1}$ $1molL^-1$ solutions EACH of $N a C l$ $NaCl$ , and $C a C l_{2}$ $CaCl_2$ ?