# The crystalline salt Na2SO4.xH2O on heating loses 55.5% of its weight the formula of crystalline salt is??? a)Na2SO4.5H2O b)Na2SO4.7H2O c)Na2SO4.2H2O d)Na2SO4.10H2O e)Na2SO4.6H2O

Sep 6, 2017

Good question, I would opt for $\left(d\right)$...

#### Explanation:

We gots $N {a}_{2} S {O}_{4} \cdot {\left(O {H}_{2}\right)}_{n}$ where $n$, the number of so-called waters of crystallization are to be determined.....

We write the dehydration reaction as follows....

$N {a}_{2} S {O}_{4} \cdot {\left(O {H}_{2}\right)}_{n} + \Delta \rightarrow N {a}_{2} S {O}_{4} + n {H}_{2} O$

We assume we start with a $100 \cdot g$ mass of sodium sulfate hydrate.....of which of course a $55.5 \cdot g$ mass are the waters of solvation, and $45.5 \cdot g$ are the $N {a}_{2} S {O}_{4}$.

$\text{Moles of sulfate,}$ $= \frac{45.5 \cdot g}{142.04 \cdot g \cdot m o {l}^{-} 1} = 0.317 \cdot m o l$ $\left(i\right)$

$\text{Moles of waters,}$ $= \frac{54.5 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 3.03 \cdot m o l$ $\left(i i\right)$

We divide $\left(i i\right)$ by $\left(i\right)$ to get the numbers of waters of crystallization, i.e. there are 10 moles of water per mole of sodium sulfate (or near enuff, I rounded up to 10 from 9.5).

And thus we got $N {a}_{2} S {O}_{4} \cdot 10 {H}_{2} O$, the so-called $\text{decahydrate}$.

Anhydrous sodium sulfate is often used as a preliminary drying agent with which to treat wet solvents......