Question

On heating $1.763g$ of hydrated $BaC{l_2}$ to dryness, $1.505g$ of anhydrous salt remained. What is the formula of hydrate?
$A)$ $BaC{l_2}.{H_2}O$
$B)$ $BaC{l_2}.3{H_2}O$
$C)$ $BaC{l_2}.5{H_2}O$
$D)$ $BaC{l_2}.2{H_2}O$

Answer 1

Hydrous compounds or hydrates are composed of water molecules in their structure as a constituent. Anhydrous compounds or anhydrates have no water molecules in the chemical structure.

Complete answer:
In the question it is given that;
Hydrated $BaC{l_2}$ ( $1.763g$ ) is heated to give anhydrous $BaC{l_2}$ ( $1.505g$ ). We have to find the formula of hydrate.
Suppose there are $'n'$ number of water molecules in the hydrated form.
$BaC{l_2}.n{H_2}O\xrightarrow{\Delta }BaC{l_2} + n{H_2}O$
Mass of Hydrated form = $1.763g$
Mass of Anhydrous form = $1.505g$
We can find the mass of water molecules in the product side by subtracting the mass of both hydrous and anhydrous (law of conservation of mass) = $1.763 - 1.505$
Mass of water = $0.258g$ (in the product side)
We can find the number of water molecules (n) by equating the moles of hydrated (product side) and Anhydrous $BaC{l_2}$ (Reactant side). Let’s find the moles of hydrated form first;
Molar mass of $BaC{l_2}$ = $208.23gmo{l^{ - 1}}$
Molar mass of ${H_2}O$ = $18gmo{l^{ - 1}}$
Moles of Hydrated form $\left( {BaC{l_2}.n{H_2}O} \right)$ = $\dfrac{{Mass}}{{Mol.mass}}$ = $\dfrac{{1.763}}{{208.23 + 18n}}$ moles
Now, moles of Anhydrous form;
Moles of $BaC{l_2}$ = $\dfrac{{1.505}}{{208.23}}$
= 0.0072 moles
We have got the moles of both hydrated and anhydrous form now we can equate them;
Moles of reactant ( $BaC{l_2}.n{H_2}O$ ) = moles of product ( $BaC{l_2}$ )
$\dfrac{{1.763}}{{208.23 + 18n}} = 0.0072$
On solving we’ll get;
$n = 2$ (approx.)
Thus, the formula of hydrate becomes;
$BaC{l_2}.2{H_2}O$
Option D is correct.

Note :
We found the number of water molecules by equating the moles of reactants and products. It can also be observed that hydrated salts can be converted to anhydrous salts by simply heating them. In many cases hydrated are coordination compounds.