InterviewSolution
Saved Bookmarks
| 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)` |
|