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 the solution was -0.53^(@)C. Assuming that the complex is formed according to the equation Hg(CN)_(2)+mCN^(-) to Hg(CN)_(m+2)^(m-) and also Hg(CN)_(2) is the limiting reactant, calculate m .

The freezing point of an aqueous solution of KCN containing 0.1892 mol

1.	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 the solution was -0.53^(@)C. Assuming that the complex is formed according to the equation Hg(CN)_(2)+mCN^(-) to Hg(CN)_(m+2)^(m-) and also Hg(CN)_(2) is the limiting reactant, calculate m .
Answer» Solution :`K_(F)=(DeltaT_(f))/(m)=(0.704)/(2xx0.1892)` (KCN DISSOCIATES COMPLETELY) `=1.86` TOTAL molality after the addition of `Hg(CN)_(2)` `=` molality of `K^(+)+` molality of `CN^(-)``+` molality of Hg `(CN)_(m+2)^(m-)` `=0.1892+(0.1892-0.095m)+0.095` `=(0.4734-0.095m)` Now, `K_(f)=(DeltaT_(f))/(m)` `1.86=(0.53)/(0.4734-0.095m)` `m=1.984` or `m=2`