Question

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 1

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`.