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. `[Co(NH_(3))_(6)]Cl_(3)`B. `[CoCl_(2)(NH_(3))_(4)]Cl`C. `[CoCl_(3)(NH_(3))_(3)]`D. `[CoCl(NH_(3))_(5)]Cl_(2)`

Answer» Correct Answer - A
Mole of `CoCl_(3).6NH_(3)=(2.675)/(267.5)=0.01mol`
`AgNO_(3)(aq)+Cl^(-)(aq)rarrAgCldownarrow(white)+NO_(3)^(-)(aq)`
mol e of AgCl=`(4.78)/(143.5)=0.03mol`
`0.01 mol CoCl_(3).6NH_(3)gives=0.03mol AgCl`
`:.1 mol CoCl_(3).6NH_(3)`ionised to give=`3 molCl^(-)`
Hence the formula of compound is `[CO(NH_(3))_(6)]Cl_(3)`


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