A solution containing 2.675 g of CoCl_3 cdot | vedclass

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(B) $1$. Calculate the moles of the complex: $n(	ext{complex}) = \frac{2.675 \ g}{267.5 \ g \ mol^{-1}} = 0.01 \ mol$. $2$. Calculate the moles of $A

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$[Co(NH_3)_6]Cl_3$

Vedclass · Answer

(B) $1$. Calculate the moles of the complex: $n(	ext{complex}) = \frac{2.675 \ g}{267.5 \ g \ mol^{-1}} = 0.01 \ mol$. $2$. Calculate the moles of $AgCl$ precipitated: $n(AgCl) = \frac{4.78 \ g}{143.5 \ g \ mol^{-1}} \approx 0.0333 \ mol$. $3$. Determine the ratio of $AgCl$ to complex: $\frac{n(AgCl)}{n(	ext{complex})} = \frac{0.0333}{0.01} \approx 3$. $4$. Since $3 \ mol$ of $Cl^-$ ions are released per mole of complex,all $3 \ Cl^-$ ions must be outside the coordination sphere. $5$. Therefore,the formula is $[Co(NH_3)_6]Cl_3$.

List-$I$ (Conductivity)	List-$II$ (Formula)
$A. \ 229$	$i. \ [Pt(NH_3)_3Cl_3]Cl$
$B. \ 97$	$ii. \ [Pt(NH_3)_4Cl_2]Cl_2$
$C. \ 404$	$iii. \ [Pt(NH_3)_5Cl]Cl_3$
$D. \ 523$	$iv. \ [Pt(NH_3)_6]Cl_4$

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