A) \[[Co{{({{H}_{2}}O)}_{4}}C{{l}_{2}}]\,Cl.2{{H}_{2}}O\]
B) \[[Co\,{{({{H}_{2}}O)}_{3}}C{{l}_{3}}]\cdot 3{{H}_{2}}O\]
C) \[[Co\,{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}\]
D) \[[Co\,{{({{H}_{2}}O)}_{5}}Cl]\,C{{l}_{2}}\cdot {{H}_{2}}O\]
Correct Answer: D
Solution :
| Molarity (M) \[\text{=}\frac{\text{Number}\,\text{of}\,\text{moles}\,\text{of}\,\text{solute}}{\text{Volume}\,\text{of}\,\text{slution}\,\text{(in}\,\text{L)}}\] |
| \[\therefore \]Number of moles of complex |
| \[=\frac{\text{Molarity}\times \text{volume}\,(\text{in}\,\text{mL})}{1000}\]\[=\frac{0.1\times 100}{1000}=0.01\,mol\] |
| Number of moles of ions precipitate \[=\frac{1.2\times {{10}^{22}}}{6.02\times {{10}^{23}}}\]\[=0.02\,mol\] |
| Number of \[C{{l}^{-}}\]present in ionization sphere |
| \[\text{=}\frac{\text{Number}\,\text{of}\,\text{moles}\,\text{of}\,\text{ions}\,\text{precipitated}}{\text{Number}\,\text{of}\,\text{moles}\,\text{of}\,\text{complex}}\]\[=\frac{0.02}{0.01}=2\] |
| \[\therefore \]\[2\,C{{l}^{-}}\]are present outside the square brackets, i.e. in iomsation sphere. Thus, the formula of complex is |
| \[[Co{{({{H}_{2}}O)}_{5}}Cl]C{{l}_{2}}\cdot {{H}_{2}}O\] |
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