A) \[AgN{{O}_{3}};1.2\times {{10}^{22}}\]
B) \[[Co{{({{H}_{2}}O)}_{3}}C{{l}_{3}}].3{{H}_{2}}O\]
C) \[[Co{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}\]
D) \[[Co{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}.{{H}_{2}}O\]
Correct Answer: D
Solution :
\[\text{Moles}\,\text{of}\,\text{complex}\,\text{=}\,\frac{\text{Molarity}\,\,\text{ }\!\!\times\!\!\text{ }\,\,\text{volume}\,\text{(ml)}}{\text{1000}}\] \[=\frac{100\times 0.1}{1000}=0.01\,mole\] Moles of ions precipitated with excess of \[AgN{{O}_{3}}=\frac{1.2\times {{10}^{22}}}{6.02\times {{10}^{23}}}\] \[=0.02\,\text{moles}\] Number of \[\text{C}{{\text{l}}^{-}}\]present in ionization sphere = \[\frac{\text{Mole}\,\text{of}\,\text{ion}\,\text{precipitated}\,\text{with}\,\text{exess}\,\text{AgN}{{\text{O}}_{\text{3}}}}{\text{Mole}\,\text{of}\,\text{complex}}=\frac{0.02}{0.01}=2\] It means \[\text{2C}{{\text{l}}^{-}}\]ions present in ionization sphere \[\therefore \]complex is \[\text{ }\!\![\!\!\text{ Co(}{{\text{H}}_{\text{2}}}\text{O}{{\text{)}}_{\text{5}}}\text{Cl }\!\!]\!\!\text{ C}{{\text{l}}_{\text{2}}}\text{.}{{\text{H}}_{\text{2}}}\text{O}\]
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