What is the proper way to prepare a solution from hydrate?

Question

I am following a protocol that requires me to prepare $10~ mathrm{mL}$ $mathrm{1~M}~ ce{Ca(NO3)2}$. However, there is only $ce{Ca(NO3)2*4H2O}$ in the lab.

Originally I thought that all I have to do to figure how much $ce{Ca(NO3)2}$ to weigh is

and then just add  $10~ mathrm{mL}$  water. However, that might dilute the solution and make the concentration less than $mathrm{1~M}$ as there is already water in the hydrate.

So then I found a protocol and made new calculations as followed:

How can I verify that I got $mathrm{1~M}$ $ce{Ca(NO3)2}$ as opposed to $mathrm{1~M}$ $ce{Ca(NO3)2*4H2O}$?

Which protocol is correct?

A.K. · Answer

Originally I thought that all I have to do to figure how much $ce{Ca(NO3)2}$ to weigh is

and then just add  $10~ mathrm{mL}$  water.

To properly make an aqueous solution of a specified concentration you should dissolve the desired amount of material in about $80%$ ($8~ mathrm{mL}$) the desired volume of water. Then dilute to the final solution volume (add water until the solution volume is $10~ mathrm{mL}$).

How can I verify that I got $mathrm{1~M}$ $ce{Ca(NO3)2}$ as opposed to $mathrm{1~M}$ $ce{Ca(NO3)2*4H2O}$?

Hydrates are for the solid state.  You do not have "hydrates" in the aqueous phase.  Therefore the two solutions are identical and indistinguishable.

What is the proper way to prepare a solution from hydrate?

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