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(D) The number of moles of $CoCl_3 \cdot 6H_2O$ is $0.1 \ M 	imes 0.1 \ L = 0.01 \ mol$.The number of moles of $AgCl$ precipitated is $\frac{1.2 	im

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$[Co(H_2O)_5 Cl]Cl_2 \cdot H_2O$

Vedclass · Answer

(D) The number of moles of $CoCl_3 \cdot 6H_2O$ is $0.1 \ M 	imes 0.1 \ L = 0.01 \ mol$.The number of moles of $AgCl$ precipitated is $\frac{1.2 	imes 10^{22}}{6.022 	imes 10^{23}} \approx 0.02 \ mol$.Since $0.01 \ mol$ of complex produces $0.02 \ mol$ of $AgCl$,$1 \ mol$ of complex produces $2 \ mol$ of $AgCl$.This indicates that there are $2$ chloride ions outside the coordination sphere.Therefore,the complex is $[Co(H_2O)_5 Cl]Cl_2 \cdot H_2O$.

On treatment of $100 \ mL$ of $0.1 \ M$ solution of $CoCl_3 \cdot 6H_2O$ with excess $AgNO_3$,$1.2 \times 10^{22}$ ions are precipitated. The complex is:

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