Question

On heating 1.763 g of hydrated BaCl, to dryness,
1.505 g of anhydrous salt remained. Hence the
formula of the hydrate is (Atomic weight of
Ba = 137)
(1) BaCl2.1/2H2O (2) BaCl2.H2O
(3) BaCl2.2H20 (4) BaCl2.5H2O​

Answer 1

Answer: Option 3

Explanation:

Mass of water lost in heating = 1.763 - 1.505 = 0.258

Moles of water we have = 0.258/18 = 0.014

Molar mass of BaCl2 = 137 + 35.5*2 = 208g/mol

Moles of anhydrous BaCl2 = 1.505/208 = 0.007 moles

Now ratio of moles of anhydrous BaCl2 and H2O = 0.007:0.014 = 1:2

Hence the formula of hydrous barium chloride = BaCl2.2H2O