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)

Explanation

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