A sample of crystalline $\mathrm{Ba}(\mathrm{OH}{)}_2·{\mathrm{XH}}_2\mathrm{O}$ weight $1.578gm$ was dissolved in water. The solution required $40 ml$ of $0.25\mathrm{N} {\mathrm{HNO}}_3$ for complete relation. Determine the number of molecules of water of crystallization in the base.

 Balance chemical reaction:

$Ba(OH{)}_2+2{HNO}_3⟶Ba{\left({NO}_3\right)}_2+H_{2}O$

According to the law of equivalence

Number of equivalence of $Ba(OH{)}_2$ = Volume$\times$Normality = 0.04 $\times$0.25 = 0.01 eq

Moles of $Ba(OH{)}_2$ = 0.001/2 = 0.005 moles

Weight of $Ba(OH{)}_2$ = moles$\times$molar mass = 0.005$\times$171.3 = 0.8565 gm

Wight of water = w – weight of $Ba(OH{)}_2$ = 1.578 - 0.8665 = 0.7215 gm

Moles of water = weight of water / molar mass of water = 0.7215 / 18

Molecules of water = mole of water $mole of Ba(OH{)}_2=\frac{\frac{0.7215}{18}}{0.005}=8.01$ ≅ 8%

Therefore, number of molecules of water is 8.