A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is : (atomic mass, Ba = 137 amu, Cl = 35.5 amu)

Options

A. `BaCl_(2), 4H_(2)O`
B. `BaCl_(2), 3H_(2)O`
C. `BaCl_(2), H_(2)O`
D. `BaCl_(2), 2H_(2)O`

Answer

D

Explanation

Weight of hydrated `BaCl_(2)=61 g`
Weight of anhydrous `BaCl_(2)=52 g`
Loss in mass = 9 g
Assuming `BaCl_(2).xH_(2)O` as hydrate
Mass of `H_(2)O=9g`
Moles of `H_(2)O=(9)/(18)=0.5`
Grass molecular let of `BaCl_(2)=208`
% of `H_(2)O` in this hydrated `BaCl_(2)=(9)/(61) xx 100=14.75%`
`=(18x)/(208+18x) xx 100` on solving x = 2
This percentage is present in `BaCl_(2).2H_(2)O`