50 , mL of 0.2 , M solution of a compound wit | vedclass

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(B) Number of millimoles of the complex $= 50 \, mL 	imes 0.2 \, M = 10 \, 	ext{mmol}$.Number of millimoles of $AgCl$ formed $= \frac{1.435 \, g}{14

Vedclass · Accepted Answer

$[Co(NH_3)_4Cl_2]Cl$

Vedclass · Answer

(B) Number of millimoles of the complex $= 50 \, mL 	imes 0.2 \, M = 10 \, 	ext{mmol}$.Number of millimoles of $AgCl$ formed $= \frac{1.435 \, g}{143.5 \, g/mol} = 0.01 \, mol = 10 \, 	ext{mmol}$.Since $10 \, 	ext{mmol}$ of the complex yields $10 \, 	ext{mmol}$ of $AgCl$,each mole of the complex releases $1$ mole of ionizable $Cl^-$ ions.This indicates the presence of one $Cl^-$ ion outside the coordination sphere.Therefore,the formula is $[Co(NH_3)_4Cl_2]Cl$.

$50 \, mL$ of $0.2 \, M$ solution of a compound with empirical formula $CoCl_3 \cdot 4NH_3$ on treatment with excess of $AgNO_{3(aq)}$ yields $1.435 \, g$ of $AgCl$. Ammonia is not removed by treatment with concentrated $H_2SO_4$. The formula of the compound is:

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