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The freezing point of an aqueous solutio...

The freezing point of an aqueous solution of KCN containing `0.1892" mole"//"kg" H_(2)O` was `-0.704^(@)C`. On adding 0.095 mole of `Hg(CN)_(2)`, the freezing point of solution was `-0.53^(@)C`. Assuming that complex is formed according to the reaction
`Hg(CN)_(2)+xCN^(-) to Hg(CN)_(x+2)^(x-)`
and also `Hg(CN)_(2)` is limiting reagent, find x.

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The correct Answer is:
`pi//2^(n-1)`
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Freezing point of 0.2 M KCN solution is -0.7^(@)C . On adding 0.1 mole of Hg(CN)_(2) to one litre of the 0.2 M KCN solution, the freezing point of the solution becomes -0.53^(@)C due to the reaction Hg(CN)_(2)+mcN^(-)toHg(CN)_(m+2)^(m-) . What is the value of m assuming molality = molarity?

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