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A solution containing 2.665 g of CrCl_3.6H_(2)O is passed through a cation exchanger. The chloride ions obtained in solution react with AgNO_3 and 2.87g of AgCl_3 is precipitated. Determine the structure of the complex. ( Cr = 52 , Cl = 35.5 , Ag = 108 , N = 14 ) |
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Answer» Solution :Amount of `CrCl_3.6H_(2)O` REACTED = 2.665 g Molar mass of the complex = `(52 + 3 xx 35.5 + 6 xx 18)` = `266.5 g mol^(-1)` No of moles of the complex reacted = `(2.665 g)/ (266.5 g mol^(-1)` = 0.01 mol Amount of AGCL formed = 2.87 g Molar mass of AgCl = `(108 + 35.5)` = `143.5 g mol^(-1)` Noof moles of AgCl formed = `(2.87 g ) / (143.5 g mol^(-1)` = 0.02 mol 0.01 mole of the complex gives AgCl = 0.02 mol 1 mole of the complex gives AgCl = 2 mol The number of free `CL^(-)` ions = 2 C.N of Cr in the complex = 6 `therefore` the complex may be represented as : `[CrClH_(2)O_(5)]Cl_(2).H_(2)O` : pentaaquachloridochromium (III) chloride monohydrate. |
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