on heating 1.763 gram of hydrated BaCl2 to dryness 1.505 gram of anhyd

1.	on heating 1.763 gram of hydrated BaCl2 to dryness 1.505 gram of anhydrous salt remained. Hence the formula of hydrated salt will be?? (atomic no of Ba = 137)
Answer» Mass of hydrated BACL2 = 1.763g Mass of DRY BaCl2 = 1.505 g Mass of Water = 1.763 - 1.505 = 0.258g Number of moles of water = 0.25818=0.014 Number of moles of BaCl2=1.505208=0.007 Ratio of moles of BaCl2 to water = 0.007 : 0.014 = 1:2 ∴ FORMULA of the hydrated salt = BaCl2.2H2O

on heating 1.763 gram of hydrated BaCl2 to dryness 1.505 gram of anhydrous salt remained. Hence the formula of hydrated salt will be?? (atomic no of Ba = 137)

Answer»

Mass of hydrated BACL2 = 1.763g

Mass of DRY BaCl2 = 1.505 g

Mass of Water = 1.763 - 1.505 = 0.258g

Number of moles of water = 0.25818=0.014

Number of moles of BaCl2=1.505208=0.007

Ratio of moles of BaCl2 to water = 0.007 : 0.014 = 1:2

∴ FORMULA of the hydrated salt = BaCl2.2H2O