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Say I have a hydrate, and I am going to dissolve it in water. Does the water in the hydrate affect the final concentration?

Here's an example: If I have $\pu{249.69 g}$ of $\ce{CuSO4·5H2O}$, and dissolve it in $\pu{200 mL}$ of water, will the concentration be $\pu{5 M}$?

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    $\begingroup$ Of course it does. The water from crystallohydrate is still water. $\endgroup$ – permeakra Jan 5 '15 at 0:28
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Of course, the water in the hydrate affects the final concentration.

  • Molar mass of $\ce{CuSO4.5H2O}$ is $\pu{249.5 g/mol}.$
  • Amount of substance is $$n = \frac{345.55}{249.5} = \pu{1.385 mol}$$
  • Concentration: $$C = \frac{1.385}{0.2} = \pu{6.925 mol/L}$$

But, this is only an illustrative example. It's not real. The solubility of $\ce{CuSO4.5H2O}$ is $\pu{320 g/L}$ at $\pu{20 °C}.$

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Firstly, copper would have to be in the 4+ oxidation state for there to be $\ce{Cu(SO4)2.5H2O}$, so this isn't a realistic question.

But, no it won't be 5M, because you need to know the final volume of the solution to calculate molarity. If you add 345.55 grams of just about anything I can think of to 200 mL of water, the volume will increase substantially. So not only will the 90 grams of hydrate water add to the volume of the final solution, generally you need to consider how all added substances contribute the final volume.

In some cases, such as magnisium sulfate, a small amount added to water can actually decrease the volume.

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Just to be clear, I don't think you can actually make a 5M solution of copper sulfate, but I'll ignore solubility for this answer.

When you make a solution of a given concentration from a solid you generally dissolve the solute (your copper sulfate pentahydrate) in a relatively small amount of water and then dilute it to the final volume. Here, you would dissolve your 249g (1 mole) of copper sulfate pentahydrate in sufficient water to dissolve it but less than 200mL total, then carefully add enough water to reach a solution volume of 200mL, yielding a 5M solution.

The water of hydration would indeed become part of the solution. If instead you began this process with anhydrous copper sulfate and used 1 mole of that (159.61g) then you would dissolve it in enough water to make the solution but less than 200mL total then carefully add enough to reach your final solution volume and concentration, 200mL at 5M.

The difference here would be that you would have to add 90g of additional water to prepare a solution from the anhydrous compound as compared to the pentahydrate.

If you prepare a solution by adding a fixed mass of solute to a fixed volume of water, you are never going to make the desired concentration and volume.

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I realize this may not be what you are asking, but this is worth pointing out. The most common source for error in solution concentration comes from neglecting to account for the mass of the hydrate. If you weight out an amount of a hydrate, and use the molecular mass of the anhydrous, your final concentration will be erroneous. For example, 5g of $\ce{CuSO4 \dot{} 5 H2O}$ will not yield the same as 5g $\ce{CuSO4}$ (anhydrous). When diluted to a volume of 1 liter:

$$\frac{\ce{5g\,CuSO4.5H2O}}{1} \frac{\ce{1mol\,CuSO4.5H2O}}{249.685\ce{g}}\frac{1}{\ce{1L}}=0.02\ce{M\,CuSO4}$$

$${\frac{\ce{5g\,CuSO4}}{1} \frac{\ce{1mol\,CuSO4}}{159.6086\ce{g}}\frac{1}{\ce{1L}}=0.03\ce{M\,CuSO4}}$$

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The molar mass of copper (ii) sulfate penta-hydrate is $\pu{249.69 g/mol}$ and that for anhydrous copper(ii) sulfate is $\pu{159.61 g/mol}$. Thus, in $1$ mole of the hydrated substance there are $\pu{159.61 g}$ of copper (ii) sulfate and $5 \times \pu{18.00 g}$ (i.e. $\pu{90.00 g}$) of molecules of water of crystallization.
$\pu{5M}$ copper (ii) sulfate solution means $5$ moles of the substance is contained in $1$ liter of solution. The mass of copper (ii) sulfate in $\pu{5M}$ of the substance is $5 \times \pu{159.61 g/L} = \pu{798.05 g/L}$.
This means that to prepare a $\pu{5 M}$ copper sulfate solution, weigh out $\pu{798.05 g}$ of the anhydrous substance and dissolve in $\pu{1 L}$ volumetric flask with distilled or de-ionized water, making up to volume by filling up to the thin ring mark at neck of flask.This can be scaled down by a factor of $5$ by dissolving $\pu{159.61 g}$ of the substance in $\pu{0.2 L}$ ($\pu{200 mL}$) solution, using a $\pu{200 mL}$ volumetric flask.

Now if the penta-hydrate is to be used to achieve a concentration of $\pu{5 M}$ copper sulfate solution, from the previous we see that $\pu{159.61 g}$ copper sulfate is contained in $\pu{249.69 g}$ of the hydrated copper sulfate. Therefore, $\pu{798.05 g}$ of copper sulfate is contained in $(249.69/159.61) \times \pu{798.05 g} = \pu{1248.45 g}$ of the pent-hydrate, i.e. dissolve $\pu{1248.45 g}$ of the hydrated salt in $\pu{1L}$.

Or if a smaller quantity is desired, say $\pu{200 mL}$ solution: weigh $(249.69/159.61) \times \pu{159.61 g} = \pu{249.69 g}$ of the penta-hydrate and dissolve in $\pu{200 mL}$ solution. Care must be taken to add water in small quantities so as not to exceed the ring graduation at neck of flask and the lower meniscus of solution should align with the mark.

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