1,L, 0.02,M solution of [Co(NH_3)_5SO_4]Br is | vedclass

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(B) The initial moles of $[Co(NH_3)_5SO_4]Br$ are $1\,L 	imes 0.02\,M = 0.02\,mol$.The initial moles of $[Co(NH_3)_5Br]SO_4$ are $1\,L 	imes 0.02\,M

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$0.01, 0.01$

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(B) The initial moles of $[Co(NH_3)_5SO_4]Br$ are $1\,L 	imes 0.02\,M = 0.02\,mol$.The initial moles of $[Co(NH_3)_5Br]SO_4$ are $1\,L 	imes 0.02\,M = 0.02\,mol$.The total volume of the mixture is $2\,L$.When the mixture is divided into two equal parts $(X)$,each part has a volume of $1\,L$.In $1\,L$ of solution $(X)$,the moles of each complex are halved: $0.01\,mol$ of $[Co(NH_3)_5SO_4]Br$ and $0.01\,mol$ of $[Co(NH_3)_5Br]SO_4$.The complex $[Co(NH_3)_5SO_4]Br$ dissociates to give $Br^-$ ions,and $[Co(NH_3)_5Br]SO_4$ dissociates to give $SO_4^{2-}$ ions.Moles of $Br^-$ in $1\,L$ of $(X) = 0.01\,mol$.Moles of $SO_4^{2-}$ in $1\,L$ of $(X) = 0.01\,mol$.Reaction with $AgNO_3$: $Br^- + AgNO_3 ightarrow AgBr(Y) + NO_3^-$. Moles of $Y (AgBr) = 0.01\,mol$.Reaction with $BaCl_2$: $SO_4^{2-} + BaCl_2 ightarrow BaSO_4(Z) + 2Cl^-$. Moles of $Z (BaSO_4) = 0.01\,mol$.

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