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)6Cl3]
- C[Co(NH3)5Cl]Cl2
- D[Co(NH3)4Cl2]NO3
✓
Answer
- [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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