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?

1 Answer
Apr 12, 2017

Answer:

Well, at a guess, you are using #Na_2S_2O_3*5H_2O#........

Explanation:

#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.