Pls solve it.

Asked by pb_ckt | 16th May, 2019, 03:45: PM

Expert Answer:

Given:
 
Mass of hydrated BaCl2 = 1.763 gm
 
Mass of anhydrous BaCl2 = 1.505 gm
 
We know,
 
space space space space BaCl subscript 2. nH subscript 2 straight O space space space space rightwards arrow with blank on top space space space BaCl subscript 2 space space plus space nH subscript 2 straight O space
open parentheses 208 space plus space straight n cross times 18 close parentheses space straight g space space space space space space space space space 208 space straight g
 
 
1.763 gm hydrated BaCl2 gives 1.505 gm of anhydrous BaCl2 
 
We know,
 
No. of moles equals space fraction numerator Weight over denominator Molecular space weight end fraction
 
fraction numerator 208 plus 18 straight n over denominator 1.763 end fraction space equals space fraction numerator 208 over denominator 1.505 end fraction

straight n space equals space fraction numerator open parentheses 1.763 cross times 208 close parentheses space minus space open parentheses 208 cross times 1.505 close parentheses over denominator 18 cross times 1.505 end fraction

space space space space equals fraction numerator space 208 cross times 0.258 over denominator 18 cross times 1.505 end fraction

space space space equals space 1.98

space space space almost equal to space 2
Hence the formula of the hydrate is BaCl2. 2H2O

Answered by Varsha | 16th May, 2019, 07:16: PM