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  • On treatment of 100 ml of 0.1M solution of the complex CrCl3.6H2O with excess of AgNo3 ,4.305g of AgCl was obtained. This complex is

On treatment of 100 ml of 0.1M solution of the complex CrCl3.6H2O with excess of AgNo3 ,4.305g of AgCl was obtained. This complex is 

Answers (1)

Moles space of space complex space equals space fraction numerator Molarity space straight x space volume space left parenthesis ml right parenthesis over denominator 1000 end fraction space equals space fraction numerator 100 space straight x space 0.1 over denominator 1000 end fraction space equals space 0.01 space mole Moles space of space ions space precipitated space with space excess space of space AgNO subscript 3 space equals space fraction numerator 1.2 space straight x space 10 to the power of 22 over denominator 6.02 space straight x space 10 to the power of 23 end fraction space equals space 0.02 space moles

Number of Cl- ion present in ionization sphere = 

fraction numerator Mole space of space ion space precipitated space with space exess space AgNO subscript 3 over denominator Mole space of space complex end fraction space equals space fraction numerator 0.02 over denominator 0.01 end fraction space equals space 2

Means 2Cl- ions are present in ionisation sphere

Hence the complex is [Co(H2O)5Cl]Cl2.H2O

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Satyajeet Kumar

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