When Hg(CN)_2 is added to a solution of compl | vedclass

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(B) The reaction is $2K^+ + 2CN^- + Hg(CN)_2 	o 2K^+ + [Hg(CN)_4]^{2-}$.Initially,there are $2$ moles of $K^+$ and $2$ moles of $CN^-$ ions,totaling

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The osmotic pressure decreases.

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(B) The reaction is $2K^+ + 2CN^- + Hg(CN)_2 	o 2K^+ + [Hg(CN)_4]^{2-}$.Initially,there are $2$ moles of $K^+$ and $2$ moles of $CN^-$ ions,totaling $4$ particles per formula unit of $KCN$ (assuming complete dissociation).After the reaction,the $CN^-$ ions are consumed to form the complex $[Hg(CN)_4]^{2-}$.The total number of particles in the solution decreases because $4$ ions $(2K^+ + 2CN^-)$ are replaced by $3$ particles $(2K^+ + [Hg(CN)_4]^{2-})$.Since colligative properties like osmotic pressure are directly proportional to the number of particles in the solution,a decrease in the number of particles leads to a decrease in osmotic pressure.

When $Hg(CN)_2$ is added to a solution of completely ionized $KCN$,the following reaction occurs: $2KCN + Hg(CN)_2 \to K_2[Hg(CN)_4]$. Due to the formation of this complex,which of the following occurs?

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