A `0.2 molal` solution of `KCl` freezes at `-0.68^(@)C`. If `K_(f)` for `H_(2)O` is `1.86`, the degree of dissociation of `KCl` is
a.`85%` , b.`83%` , c. `65%` , d. `90%`

Explanation

To find the degree of dissociation of KCl in a 0.2 molal solution that freezes at -0.68°C, we can follow these steps:

### Step 1: Calculate the Depression in Freezing Point (ΔTf)
The depression in freezing point (ΔTf) can be calculated as:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]
Given that the freezing point of pure water (T_f^{pure}) is 0°C and the freezing point of the solution (T_f^{solution}) is -0.68°C, we have:
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