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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$0.01, 0.01$

Vedclass · Answer

(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 $Z (BaSO_4) = 0.01\,mol$.

Similar Questions

Sulphide ions react with sodium nitroprusside in a basic medium giving a coloured solution. In the reaction,the oxidation state of iron:

Identify the geometrical isomers of the following complexes:

$0.001 \ mol$ of $Co(NH_3)_5(NO_3)(SO_4)$ was passed through a cation exchanger. The acid coming out of it required $20 \ mL$ of $0.1 \ M$ $NaOH$ for neutralization. Hence,the complex is:

$A$ blue colouration is not obtained when:

Vedclass Test Series

Exam Paper Generator

Online Exam Module