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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 freezing point of an aqueous solution of KCN containing 0.1892 mol kg^(-1) was -0.704^(circ)C . On adding 0.45 mole of Hg(CN)_(2) , the freezing point of the solution was =0.620^(circ)C . If whole of Hg(CN)_(2) is used in complex formation according to the equation, Hg(CN)_(2)+mKCN rarrK_(m)[Hg(CN)_(m+2)] what is the formula of the complex ? Assume [Hg(CN)_(m+2)]^(m-) is not ionised and the complex molecule is 100% ionised. (K_(f)(H_(2)O) is 1.86 kg mol^(-1) K .)

The freezing point of an aqueous solution of KCN containing 0.1892 mol Kg^(-1) , the freezing point of the solution was found to be -0.530^(@)C . If the complex formation takes place according to the following equation: Hg(CN)_(2) , the freezing point of the solution was found to be -0.530^(@)C . If the complex formation takes place according to the following equation: Hg(CN)_2 +nKCN hArr K_n[Hg (CN)_(n+2)] what is the formula of the complex? [ K_(f)(H_(2)O) is 1.86 K kg mol^(-1) ]

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