A) \[\left[ Co{{\left( {{H}_{2}}O \right)}_{6}} \right]C{{l}_{3}}\]
B) \[\left[ Co\left( {{H}_{2}}O \right)5Cl \right]C{{l}_{2}}.{{H}_{2}}O\]
C) \[\left[ Co{{\left( {{H}_{2}}O \right)}_{3}}C{{l}_{3}} \right].3{{H}_{2}}O\]
D) \[\left[ Co{{\left( {{H}_{2}}O \right)}_{4}}C{{l}_{2}} \right]Cl.2{{H}_{2}}O\]
Correct Answer: A
Solution :
As mol of \[AgCl\,\,=\,\,\frac{4.305}{143.5}\,\,=\,\,0.03\] = mol of \[C{{l}^{-}}\] given by complex mol of the complex \[=\text{ }100\,\,\times \,\,{{10}^{-}}^{3}\times 1\,\,=\,\,0.01\] \[\left[ Co{{\left( {{H}_{2}}O \right)}_{6}} \right]C{{l}_{3}}\to \,\,\,{{\left[ Co{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{+}}^{3}+\,\,3C{{l}^{-}}\] 0.01 mole 0.03 mole
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