`1g` of a monobasic acid dissolved in `200g` of water lowers the freezing point by `0.186^(@)C`. On the other hand when `1g` of the same acid is dissolved in water so as to make the solution `200mL`, this solution requires `125mL` of `0.1 N NaOH` for complete neutralization. Calculate `%` dissociation of acid ? `(K_(f)=1.86(K-kg)/("mol"))`

Explanation

To solve the problem, we need to find the percentage dissociation of a monobasic acid (HA) based on the provided data. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the molality of the solution
We know that the depression in freezing point (ΔTf) is given by the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Where:
- \( \Delta T_f = 0.186 \, ^\circ C \)
- \( K_f = 1.86 \, \text{(K kg/mol)} \)
- \( m \) is the molality
- \( i \) is the van't Hoff factor

Rearranging the formula gives:
\[ m \cdot i = \frac{\Delta T_f}{K_f} = \frac{0.186}{1.86} = 0.1 \]

### Step 2: Calculate the number of moles of the acid
When 1 g of the acid is dissolved in 200 mL of solution, it requires 125 mL of 0.1 N NaOH for complete neutralization.

First, calculate the number of moles of NaOH used:
\[ \text{Number of moles of NaOH} = \text{Normality} \times \text{Volume in L} \]
\[ = 0.1 \, \text{N} \times \frac{125 \, \text{mL}}{1000} = 0.0125 \, \text{moles} \]

Since the acid is monobasic (HA), it will react with NaOH in a 1:1 ratio. Therefore, the number of moles of HA is also:
\[ \text{Number of moles of HA} = 0.0125 \, \text{moles} \]

### Step 3: Calculate the molality of the acid
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Given that the solvent (water) is 200 g, we convert this to kg:
\[ \text{Weight of solvent} = 200 \, \text{g} = 0.2 \, \text{kg} \]
Now, we can calculate the molality:
\[ m = \frac{\text{Number of moles of HA}}{\text{Weight of solvent in kg}} = \frac{0.0125}{0.2} = 0.0625 \, \text{mol/kg} \]

### Step 4: Calculate the van't Hoff factor (i)
From Step 1, we have:
\[ m \cdot i = 0.1 \]
Substituting the value of molality:
\[ 0.0625 \cdot i = 0.1 \]
Solving for \( i \):
\[ i = \frac{0.1}{0.0625} = 1.6 \]

### Step 5: Relate the van't Hoff factor to the degree of dissociation
For a monobasic acid (HA), the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
The van't Hoff factor \( i \) can be expressed as:
\[ i = 1 + (n - 1) \cdot \alpha \]
Where \( n \) is the number of ions formed upon dissociation (which is 2 for HA) and \( \alpha \) is the degree of dissociation.

Substituting \( n = 2 \):
\[ 1.6 = 1 + (2 - 1) \cdot \alpha \]
\[ 1.6 = 1 + \alpha \]
Solving for \( \alpha \):
\[ \alpha = 1.6 - 1 = 0.6 \]

### Step 6: Calculate the percentage dissociation
The percentage dissociation is given by:
\[ \text{Percentage dissociation} = \alpha \times 100 = 0.6 \times 100 = 60\% \]

### Final Answer
The percentage dissociation of the acid is **60%**.

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