MCQ
When 0.01 mole of a cobalt complex is treated with excess silver nitrate solution, 4.305 g silver chloride is precipitated. The formula of the complex is:
  • A
    [Co(NH3​)6​]Cl3
  • B
    [Co(NH3​)6​Cl3​]
  • C
    [Co(NH3​)5​Cl]Cl2
  • D
    [Co(NH3​)4​Cl2​]NO3​

Answer

  1. [Co(NH3​)6​]Cl3

Explanation:

4.305gAgCl = $\frac{4.305}{143.5}$​ mol = 0.03mol.

As 0.01 mole of the complex gives 0.03 mole of AgCI, this shows that there are three ionisable Cl.

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