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0.7875 g of crystalline barium hydroxide...

0.7875 g of crystalline barium hydroxide is dissolved in water .For the neutralization of this solution 20 mL of N/4 `HNO_(3)` is required. How many moles of water of crystallization are present in one mole of this base ? (Given : Atomic mass Ba=137,O=16, N=14, H=1)

Text Solution

Verified by Experts

Let the molecular formula be `Ba(OH)_(2). xH_(2)O`
Mol. Mass of `Ba(OH)_(2).xH_(2)O=137.4+(2xx16)+(2xx1)+18x`
=171.4+18x
Eq. mass of `Ba(OH)_(2).xH_(2)O=(171.4+18x)/(2)`
`20 mL(N)/(4)HNO_(3)-=20mL(N)/(4)Ba(OH)_(2).xH_(2)O`
Amount of `Ba(OH)_(2).xH_(2)O=(171.4+18x))/(2xx4)xx(20)/(1000)`
`=(171.4+18x)/(400)g`
Amount of `Ba(OH)_(2).xH_(2)O=0.789 g`
Hence, `(171.4+18x)/(400)=0.789`
or `171.4+18x=0.789xx400`
`x=(144.2)/(18)=8.01~~8`
Thus, 8 g moles of water molecules are present in one g mole of the base.
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