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 .

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the freezing point depression (ΔTf)
The freezing point depression is calculated as:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]
Given that the freezing point of the pure solvent (water) is \(0^\circ C\) and the freezing point of the solution is \(-0.704^\circ C\):
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