`1.5g` of monobasic acid when dissolved in `150g` of water lowers the freezing point by `0.165^(@)C.0.5g` of the same acid when titrated, after dissolution in water, requires `37.5 mL` of `N//10` alkali. Calculate the degree of dissociation of the acid `(K_(f)` for water `= 1.86^(@)C mol^(-1))`.

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Understand the Problem
We need to calculate the degree of dissociation (α) of a monobasic acid based on the freezing point depression and titration data provided.

### Step 2: Use the Freezing Point Depression Formula
The formula for depression in freezing point is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \(\Delta T_f\) = depression in freezing point
- \(i\) = Van't Hoff factor (degree of dissociation)
- \(K_f\) = cryoscopic constant of the solvent (water in this case)
- \(m\) = molality of the solution

### Step 3: Calculate Molality (m)
Molality (m) is defined as:
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} \]
We need to find the number of moles of the solute (the acid).

### Step 4: Calculate Molar Mass of the Acid
From the titration data, we know that 0.5 g of the acid requires 37.5 mL of \(N/10\) alkali for neutralization.

First, calculate the number of equivalents of the acid:
\[ \text{Normality} = \frac{N}{10} = 0.1 \, N \]
\[ \text{Volume} = 37.5 \, mL = 0.0375 \, L \]
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume} = 0.1 \times 0.0375 = 0.00375 \, eq \]
Since the acid is monobasic, the number of moles of the acid is equal to the number of equivalents:
\[ \text{Moles of acid} = 0.00375 \, mol \]
Now, we can find the molar mass (M) of the acid:
\[ \text{Molar mass} = \frac{\text{mass of acid}}{\text{moles of acid}} = \frac{0.5 \, g}{0.00375 \, mol} = 133.33 \, g/mol \]

### Step 5: Calculate Molality
Now we can calculate the molality using the mass of the solvent (150 g = 0.150 kg):
\[ \text{Number of moles of solute} = \frac{1.5 \, g}{133.33 \, g/mol} = 0.01125 \, mol \]
\[ m = \frac{0.01125 \, mol}{0.150 \, kg} = 0.075 \, mol/kg \]

### Step 6: Substitute Values into Freezing Point Depression Formula
Now we can substitute the values into the freezing point depression formula:
\[ \Delta T_f = 0.165 \, °C \]
\[ K_f = 1.86 \, °C \cdot kg/mol \]
Substituting into the equation:
\[ 0.165 = i \cdot 1.86 \cdot 0.075 \]
\[ i = \frac{0.165}{1.86 \cdot 0.075} = \frac{0.165}{0.1395} \approx 1.182 \]

### Step 7: Calculate Degree of Dissociation (α)
Since the acid is monobasic, the Van't Hoff factor \(i\) can be expressed as:
\[ i = 1 + \alpha \]
Thus, we can find α:
\[ 1 + \alpha = 1.182 \implies \alpha = 1.182 - 1 = 0.182 \]

### Final Answer
The degree of dissociation (α) of the acid is approximately:
\[ \alpha \approx 0.182 \text{ or } 18.2\% \]