$Na_2S_2O_3*5H_2O$ 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.
