On treating 100 mL of 0.1 M aqueous solutio | vedclass

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(C) Given,Molarity of the complex $= 0.1 \ M$. Volume of the complex $= 100 \ mL$. Mass of $AgCl$ obtained $= 2.86 \ g$. Molar mass of $AgCl = 108 + 3

Vedclass · Accepted Answer

$[Cr(H_2 O)_5 Cl] Cl_2 \cdot H_2 O$

Vedclass · Answer

(C) Given,Molarity of the complex $= 0.1 \ M$. Volume of the complex $= 100 \ mL$. Mass of $AgCl$ obtained $= 2.86 \ g$. Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$. Number of moles of complex $= \frac{0.1 	imes 100}{1000} = 0.01 \ mol$. Number of moles of $AgCl$ precipitated $= \frac{2.86 \ g}{143.5 \ g/mol} \approx 0.02 \ mol$. Since $1 \ mol$ of complex gives $2 \ mol$ of $AgCl$,there must be $2$ ionizable $Cl^-$ ions outside the coordination sphere. Therefore,the complex is $[Cr(H_2 O)_5 Cl] Cl_2 \cdot H_2 O$.

On treating $100 \ mL$ of $0.1 \ M$ aqueous solution of the complex $CrCl_3 \cdot 6 H_2 O$ with excess of $AgNO_3$,$2.86 \ g$ of $AgCl$ was obtained. The complex is:

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