On heating `1.763 g` of hydrated `BaCl_(2)` to dryness, `1.505 g` of anhyrous salt remained, What is the formula of hydrate?

Answer

2

Explanation

`BaCl_(2).xH_(2)Ooverset(Delta)rarrBaCl_(2)+xH_(2)O`
Mol.wt. `(208+18x) , 208`
`because (208+18x)BaCl_(2).xH_(2)O gives = 208 g BaCl_(2)`
`therefore 1.763g BaCl_(2).xH_(2)O=(208xx1.763)/((208+18x))gBaCl_(2)`
`therefore (208xx1.763)/((208+18x))=1.505`
`x=approx2`