When 0.1 mole of CoCl_3(NH_3)_5 is treated w | vedclass

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(B) One mole of $AgNO_3$ precipitates one mole of chloride ion. In the given reaction,$0.1 \ mole$ of $CoCl_3(NH_3)_5$ reacts to give $0.2 \ mole$ of

Vedclass · Accepted Answer

$1 : 2$ electrolyte

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(B) One mole of $AgNO_3$ precipitates one mole of chloride ion. In the given reaction,$0.1 \ mole$ of $CoCl_3(NH_3)_5$ reacts to give $0.2 \ mole$ of $AgCl$,which means $2 \ mole$ of $Cl^{-}$ ions are ionizable. The complex is $[Co(NH_3)_5Cl]Cl_2$. The electrolytic solution contains $[Co(NH_3)_5Cl]^{2+}$ and $2Cl^{-}$ as constituent ions. Thus,it is a $1 : 2$ electrolyte. $[Co(NH_3)_5Cl]Cl_2 \rightarrow [Co(NH_3)_5Cl]^{2+}_{aq} + 2Cl^{-}_{aq}$. Hence,option $B$ is correct.

When $0.1 \ mole$ of $CoCl_3(NH_3)_5$ is treated with excess of $AgNO_3$,$0.2 \ mole$ of $AgCl$ are obtained. The conductivity of the solution will correspond to:

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