A sample of crystalline $Ba (OH)_{2} \cdot {XH}_{2} O$ weight $1.578 g m$ was dissolved in water. The solution required $40 m l$ of $0.25 N {HNO}_{3}$ for complete relation. Determine the number of molecules of water of crystallization in the base.

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answer is 8.

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Detailed Solution

Balance chemical reaction:

$B a (O H)_{2} + 2 {H N O}_{3} ⟶ B a {({N O}_{3})}_{2} + H_{2} O$

According to the law of equivalence

Number of equivalence of $B a (O H)_{2}$ = Volume $\times$ Normality = 0.04 $\times$ 0.25 = 0.01 eq

Moles of $B a (O H)_{2}$ = 0.001/2 = 0.005 moles

Weight of $B a (O H)_{2}$ = moles $\times$ molar mass = 0.005 $\times$ 171.3 = 0.8565 gm

Wight of water = w – weight of $B a (O H)_{2}$ = 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 $m o l e o f B a (O H)_{2} = \frac{\frac{0.7215}{18}}{0.005} = 8.01$ ≅ 8%

Therefore, number of molecules of water is 8.

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