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A solution containing 2.665 g of CrСl_(3).6H_(2)O is passed through a cation exchanger.The chloride ions obtained in solution react with AgNO_(3) and give 2.87 g of AgCI. Determine the structure of the compound. |
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Answer» Solution :`143.45 g` of AgCl contains 35.45 g Cl" ions therefore 2.879 of A` CI_(-)` will contain `(35.45xx 2.87)/(143.45)=o.709 GCL^(-)ions` `CrCl_(3).6H_(2)0` contains `n xx 35.45` g of ionizable CR ions (where n = no. of.`Cl^(-)` ions outside the coordination SPHERE). thus, 2.665 g `ClCl_(3)6H_(2)O` will contain `(NXX 35.45 xx2.665)/(266.35)` Also `nxx 35.45xx 2.665)/(266.35)= 0.709 implies napprox2` Keeping in view the octahedral geometry of the complex, its structure may be written as `[CrCI(_(2)O)_(5),Cl_(2)H_(2)O]`. |
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