Final answer:
The complex formed when 0.1M CoCl₃.6H₂O is treated with an excess of AgNO3, leading to the precipitation of 1.2×10²² ions, is [Co(H₂O)₄Cl]Cl₂ . H₂O.
Step-by-step explanation:
The question inquires about the complex formed when a 0.1M solution of CoCl₃.6H₂O is treated with an excess of AgNO3, resulting in the precipitation of 1.2×10²² ions. With the given data and considering the principles of precipitation and complex formation with AgNO3, we can deduce the number of moles of precipitated chloride ions and compare that to the initial moles of CoCl₃.6H₂O to determine the correct complex.
Since the mole ratio of Ag+ to Cl- is 1:1 in the precipitate of AgCl, and knowing that the number of moles of Ag+ from AgCl is the same as the number of Cl- ions precipitated, we use Avogadro's number to find that the total moles of precipitated Cl- ions is 1.2×10²² ions / 6.022×10²³ ions/mol = 0.002 moles. Given that the initial solution is 0.1M, this means there were 0.01 moles of CoCl₃·6H₂O (as Molarity = moles/volume in liters, and the volume is 0.1 L).
Considering the stoichiometry of complex ions and the fact not all Cl- ions in the complex will precipitate, the correct complex where two chloride ions remain bound to cobalt would be [Co(H₂O)₄Cl]Cl₂ . H₂O.