1g of a monabasic acid when dissolved in 100g of water lowers the freezing point by `0.168^(@)C`. `0.2g` of the same acid when dissolved and titrated required `15.1mL` of `N//10` alkali . Calculate the degree of dissociation of the acid. (`K_(f)` for water is `1.86`)

Explanation

To solve the problem step by step, we will follow the provided information and apply the relevant formulas.

### Step 1: Calculate the molality of the solution
We know that the freezing point depression (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = K_f \times m \]
where \( K_f \) is the cryoscopic constant of the solvent (water in this case) and \( m \) is the molality of the solution.
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