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(D) Number of moles of the given complex $= 10 \ mL 	imes 1 \ M = 10 \ mmol = 0.01 \ mol$. Number of moles of $AgCl$ formed $= \frac{4.305 \ g}{143.5

Vedclass · Accepted Answer

$[Cr(H_2O)_6]Cl_3$

Vedclass · Answer

(D) Number of moles of the given complex $= 10 \ mL 	imes 1 \ M = 10 \ mmol = 0.01 \ mol$. Number of moles of $AgCl$ formed $= \frac{4.305 \ g}{143.5 \ g/mol} = 0.03 \ mol$. Since $AgNO_3$ is in excess,all ionizable $Cl^-$ ions react to form $AgCl$. Ratio of moles of $AgCl$ to moles of complex $= \frac{0.03 \ mol}{0.01 \ mol} = 3$. This indicates that each mole of the complex releases $3 \ mol$ of $Cl^-$ ions. Therefore,the complex must be $[Cr(H_2O)_6]Cl_3$.

On treatment of $10 \ mL$ of $1 \ M$ solution of the complex $CrCl_3 \cdot 6H_2O$ with excess of $AgNO_3$,$4.305 \ g$ of $AgCl$ was obtained. The complex is

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