A) \[[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}\]
B) \[[Co{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}\]
C) \[[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Cl\]
D) \[[Co{{(N{{H}_{3}})}_{3}}C{{l}_{3}}]\]
Correct Answer: A
Solution :
[a] Mw of \[AgCl=108+35.5=143.5\text{ }g\] Moles of \[AgCl=\frac{4.305}{143.5}=0.3mole\] Mw of \[CoC{{l}_{3}}.6N{{H}_{3}}=267.5g\] Moles of complex \[=\frac{2.675}{267.5}=0.01\] mole Therefore \[0.01\] mole of complex gives \[0.03\] mole of \[AgCl\]with \[AgN{{O}_{3}}\] Thus, \[3C{{l}^{\bigcirc -}}\] ions are present as counter ions. \[\therefore \] Formula of complex is \[\underset{1mole\,\,of\,\,complex}{\mathop{[Co{{\left( N{{H}_{3}} \right)}_{6}}]C{{l}_{3}}}}\,\xrightarrow{AgN{{O}_{3}}}\underset{white\,\,ppt.}{\mathop{3AgCl}}\,\]
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