<u>Given:</u>
Mass of hydrated barium chloride = 3.364 g
Total volume of barium chloride V(total)= 250 ml
Volume taken for titration V = 10 ml
Volume of AgNO3 consumed = 46.92 ml
Concentration of AgNO3 = 0.0253 M
<u>To determine:</u>
The value of x i.e. the water of hydration in BaCl2
<u>Explanation:</u>
The net ionic equation is-
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Based on the reaction stoichiometry: Equal moles of Ag+ and Cl- combine to form AgCl
Moles of Ag+ consumed = moles of Cl- present
Moles of Ag+ = V(AgNO3) * M(AgNO3) = 0.04692 * 0.0253 = 0.00119moles
Moles of Cl- present = 0.00119 moles
Thus, 0.00119 moles of Cl- are present in 10 ml of the solution
Therefore, number of moles of Cl- in 250 ml would be-
= 0.00119 * 250 /10 = 0.02975 moles of cl-
Now:
2 moles of Cl- are present in 1 mole of BaCl2
Therefore, 0.02975 moles of Cl- correspond to- 0.02975 * 1/2 = 0.01488 moles of BaCl2
Molar mass of BaCl2 = 208.22 g/mol
Thus, mass of BaCl2 = 0.01488 moles * 208.22 g.mol-1 = 3.098 g
Mass of water of hydration = 3.364 - 3.098 = 0.266 g
# moles of water 'x' = .266/18 = 0.015 ≅ 1
Ans: Formula for hydrated barium chloride = BaCl2. 1H2O
