A six-coordinate complex with the formula CrC | vedclass

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Question

(B) The molar mass of $AgCl$ is $143.5 \ g/mol$. Given mass of $AgCl$ precipitate = $28.7 \ g$. Number of moles of $AgCl$ = $\frac{28.7 \ g}{143.5 \ g

Vedclass · Accepted Answer

$[CrCl(H_2O)_5]Cl_2 \cdot H_2O$

Vedclass · Answer

(B) The molar mass of $AgCl$ is $143.5 \ g/mol$. Given mass of $AgCl$ precipitate = $28.7 \ g$. Number of moles of $AgCl$ = $\frac{28.7 \ g}{143.5 \ g/mol} = 0.2 \ mol$. Since $0.1 \ mol$ of the complex produces $0.2 \ mol$ of $AgCl$,each mole of the complex releases $2 \ mol$ of $Cl^-$ ions. This indicates that $2 \ Cl^-$ ions are ionizable (outside the coordination sphere). Therefore,the complex is $[CrCl(H_2O)_5]Cl_2 \cdot H_2O$.

$A$ six-coordinate complex with the formula $CrCl_3 \cdot 6H_2O$ has a green color. $A$ $0.1 \ M$ solution of the complex,when treated with an excess of $AgNO_3$,yields $28.7 \ g$ of a white precipitate. What is the formula of the complex?

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