A solution is prepared from `25*g` of salt dissolved in `1*L` of water. If the salt is an hydrate of sodium thiosulfate, how many water molecules are associated with this salt?

`Na_2S_2O_3*5H_2O` is the common laboratory reagent, this so-called pentahydrate forms large sugary crystals so it is easy and convenient to handle. The formula mass of this material is `248.18*g*mol^-1` (i.e. this represents the mass of the sodium, the sulfur, the oxygen, and the hydrogen atoms, i.e. INCLUDING the water molecules). So here a mass of `248.18*g` of this stuff represents a given quantity (a mole) of such species.

Now concentration is typically defined as the quotient:

`"Concentration"` `=` `"Amount of substance (in moles)"/"Volume of Solution (in litres)"`, and thus has units `mol*L^-1`.

So, given that we dissolve `25*g` of the salt in `1*L` of water (and I am making an educated(?) guess here), we have a concentration of.......

`"Concentration"=((25*g)/(248.18*g*mol^-1))/(1*L)=(0.100*mol)/(1*L)=0.100*mol*L^-1.`

OR `0.100*"N"` with respect to `Na_2S_2O_3*5H_2O`, i.e. the `"pentahydrate"`. We write `N` for `"normal"`, because formally we have a solution whose concentration with respect to `Na_2S_2O_3*5H_2O` is the same. Of course, when we dissolve this species up in aqueous (water) solution, we get `2xxNa^+` ions, and `1xxS_2O_3^(2-)` ions IN SOLUTION, as well the hydrate molecules to float round in the water solution INDEPENDENTLY.

If this is not clear, or if I have glossed over a step or made an assumption you do not follow, raise a query, and I or someone else will attempt to re-explain.