Say I have 0.72 grams of calcium chloride dihydrate. If I dissolve it in water, the 2 water molecules attached onto it dissociates just like the $\ce{Ca^{2+}}$ and $\ce{Cl-}$, right?

And if so, wouldn't it lower the concentration of the calcium chloride, as compared to making an anhydrous calcium chloride solution?

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    $\begingroup$ Instead of our answering this question, why don't you compute the difference assuming the two waters of hydration just sum into the total volume. Try this for a few different target concentrations: 0.001 M, 0.1 M, 10 M. $\endgroup$ – Zhe Jan 16 '17 at 23:52

There are two methods to dissolve the calcium chloride salt.

For the first method you'd dissolve the salt in a "small" amount of water and dilute to some particular volume. So if the solution was 0.100 molar, the amount of water would be the same regardless if you started with $\ce{CaCl2}$ or $\ce{CaCl2\cdot2H2O}$.

The other method would be to make say a 1.00 % solution by weight of the calcium chloride salt. Here it does matter. Do you want 1.00 % solution of $\ce{CaCl2}$ or a 1.00 % solution of $\ce{CaCl2\cdot2H2O}$?

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  • $\begingroup$ I think CaCl2, as I want to react it with another compound, and the hydrate will not participate in the reaction. $\endgroup$ – David Liao Jan 18 '17 at 1:29

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