# 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?

Apr 12, 2017

Well, at a guess, you are using $N {a}_{2} {S}_{2} {O}_{3} \cdot 5 {H}_{2} O$........

#### Explanation:

$N {a}_{2} {S}_{2} {O}_{3} \cdot 5 {H}_{2} O$ 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 $248.18 \cdot g \cdot m o {l}^{-} 1$ (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 $248.18 \cdot g$ of this stuff represents a given quantity (a mole) of such species.

Now concentration is typically defined as the quotient:

$\text{Concentration}$ $=$ $\text{Amount of substance (in moles)"/"Volume of Solution (in litres)}$, and thus has units $m o l \cdot {L}^{-} 1$.

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

$\text{Concentration} = \frac{\frac{25 \cdot g}{248.18 \cdot g \cdot m o {l}^{-} 1}}{1 \cdot L} = \frac{0.100 \cdot m o l}{1 \cdot L} = 0.100 \cdot m o l \cdot {L}^{-} 1.$

OR $0.100 \cdot \text{N}$ with respect to $N {a}_{2} {S}_{2} {O}_{3} \cdot 5 {H}_{2} O$, i.e. the $\text{pentahydrate}$. We write $N$ for $\text{normal}$, because formally we have a solution whose concentration with respect to $N {a}_{2} {S}_{2} {O}_{3} \cdot 5 {H}_{2} O$ is the same. Of course, when we dissolve this species up in aqueous (water) solution, we get $2 \times N {a}^{+}$ ions, and $1 \times {S}_{2} {O}_{3}^{2 -}$ 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.