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Q. On treatment of 100 ml of 0.1 M solution of the complex CrCl3.6H2O with excess of AgNO3, 4.305 g of AgCl was obtained. The complex is

NTA AbhyasNTA Abhyas 2020Coordination Compounds

Solution:

Mol of AgCl $= \frac{4 \text{.} 3 0 5}{1 4 3 \text{.} 5} = 0 \text{.} 0 3$ = mol of Cl- given by the complex.

Mol of the complex = 100 x 10-3 x 0.1 = 0.01 ;

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