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?

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Determine the change in freezing point (ΔTf)
The initial freezing point of the 0.2 M KCN solution is -0.7°C, and after adding 0.1 mole of Hg(CN)₂, the freezing point becomes -0.53°C.

\[ \Delta T_f = T_f^{initial} - T_f^{final} = -0.7°C - (-0.53°C) = -0.7 + 0.53 = -0.17°C \]
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