Question

on heating 1.763g of hydrated bacl2 to dryness,1.505 of anhydrous salt remained.hence the formula of the hydrate is (atomic weight of ba=137)

Answer 1

answer for this question is bacl2.2H2o

Answer 2

Answer: $BaCl_2.2H_2O$

Explanation:

$BaCl_2.xH_2O\rightarrow BaCl_2+xH_2O$

Molar mass of $BaCl_2$ = 137+71 =208 g/mol

According to stoichiometry:

(208+18x) g of $BaCl_2.xH_2O$ decomposes to give 208 g of $BaCl_2$

Thus 1.763 g of of $BaCl_2.xH_2O$ decomposes to give =$\frac{208}{208+18x}\times 1.763g$ of $BaCl_2$

As given :

$\frac{137}{137+18x}\times 1.763g=1.505g$

$x=2$  

Thus the value of x is 2 and formula of hydrate is $BaCl_2.2H_2O$