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On treatment of 100 ml of 0.1 M solution of CoCl_(3).6H_(2)O with axcess AgNO_(3), 1.2xx10^(22) ions are precipitated. The complex is |
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Answer» `[Co(H_(2)O)_(6)]Cl_(3)` `=100xx0.1` millimole = 10 millimole = 0.01 mole As the IONS produced are precipitated by `AgNO_(3)`, the ions produced must be `Cl^(-)` ions. Thus, 0.01 mole of the complex give `Cl^(-)` ions `=1.2xx10^(22)` `=(1.2xx10^(22))/(6.0xx10^(23))` mole = 0.02 mole `therefore` 1 mole of the complex gives `Cl^(-)` ions = 2 moles (which are precipitated as AgCl) Remembering that COORDINATION number of Co is 6 and TWO `Cl^(-)` ions are present outside the coordination sphere, the formula of the complex will be `[Co(H_(2)O)_(5)Cl]Cl_(2).H_(2)O`. |
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