A solution containing `2.675g` of `COCl_(3).6NH_(3)` (molar mass `=267.5gmol^(-1)`) is passed through a cation exchanger, The chloride ions obtined in solution were treated with excess of `AgNO_(3)` to give`4.78g` of `AgCl ("molar mass" =143.5gmol^(-1)`.The formula of the complex is (Atomic mass of `Ag=108 u`)A. (a) `[Co(NH_3)_6]Cl_3`B. (b) `[Co(NH_3)_3Cl_3]_3NH_3`C. (c) `[Co(NH_3)_4Cl_2]Cl_2NH_3`D. (d) `[Co(NH_3)_5Cl]Cl_2NH_3`

A solution containing `2.675g` of `COCl_(3).6NH_(3)` (molar mass `=267

1.	A solution containing `2.675g` of `COCl_(3).6NH_(3)` (molar mass `=267.5gmol^(-1)`) is passed through a cation exchanger, The chloride ions obtined in solution were treated with excess of `AgNO_(3)` to give`4.78g` of `AgCl ("molar mass" =143.5gmol^(-1)`.The formula of the complex is (Atomic mass of `Ag=108 u`)A. (a) `[Co(NH_3)_6]Cl_3`B. (b) `[Co(NH_3)_3Cl_3]_3NH_3`C. (c) `[Co(NH_3)_4Cl_2]Cl_2NH_3`D. (d) `[Co(NH_3)_5Cl]Cl_2NH_3`
Answer» Correct Answer - A No. of moles complex `=(2.675)/(267.5)=0.01` No. of moles of `AgCl=(4.78)/(143.5)=0.03` This shows that three `Cl^-` ions are precipitated. Thus, formula of the complex is `[Co(NH_3)_6]Cl_3`.

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