2.44 g 'w' g of the benzoic acid dissolved in 30 g of benzene shows depression in freezing equal to 2K. If the percentage association of the acid to form polymer `(C_(6)H_(5)COOH)_(n)` in the solution is 80, then find numerical value of n. (Given that `K_(f) = 5K kg mol^(-1)`, Molar mass of benzoic acid `= 122 g mol^(-1)`)

Explanation

To solve the problem step by step, we need to use the concepts of freezing point depression and the degree of association of benzoic acid in benzene.

### Step 1: Understand the given data
- Mass of benzoic acid (w) = 2.44 g
- Mass of benzene (solvent) = 30 g
- Depression in freezing point (ΔTf) = 2 K
- Kf (freezing point depression constant for benzene) = 5 K kg mol⁻¹
- Molar mass of benzoic acid = 122 g mol⁻¹
- Percentage association of benzoic acid = 80%

### Step 2: Calculate the molality (m) of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. Calculate the number of moles of benzoic acid:
\[ \text{Number of moles of benzoic acid} = \frac{\text{mass}}{\text{molar mass}} = \frac{2.44 \, \text{g}}{122 \, \text{g/mol}} = 0.0200 \, \text{mol} \]

2. Convert the mass of benzene to kilograms:
\[ \text{Mass of benzene in kg} = \frac{30 \, \text{g}}{1000} = 0.030 \, \text{kg} \]

3. Calculate molality (m):
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0200 \, \text{mol}}{0.030 \, \text{kg}} = 0.667 \, \text{mol/kg} \]

### Step 3: Use the freezing point depression formula
The formula for freezing point depression 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 association)
- \( K_f \) = freezing point depression constant
- \( m \) = molality

### Step 4: Calculate the van 't Hoff factor (i)
Since 80% of benzoic acid associates, we can calculate the van 't Hoff factor (i):
1. Let \( x \) be the degree of association. Then:
\[ x = 0.8 \quad (\text{since 80% associates}) \]
2. The number of moles associated = \( \frac{x}{2} \) (since two molecules associate to form one dimer).
3. The number of moles that remain unassociated = \( 1 - x = 0.2 \).

Thus, the van 't Hoff factor (i) can be calculated as:
\[ i = (1 - x) + \frac{x}{2} = 0.2 + \frac{0.8}{2} = 0.2 + 0.4 = 0.6 \]

### Step 5: Substitute values into the freezing point depression equation
Now we can substitute the values into the freezing point depression formula:
\[ 2 = 0.6 \cdot 5 \cdot m \]
\[ 2 = 3 \cdot m \]
\[ m = \frac{2}{3} \approx 0.667 \, \text{mol/kg} \]

### Step 6: Calculate the number of moles of solute (W)
Using the molality we calculated:
\[ m = \frac{W}{\text{mass of solvent in kg}} \Rightarrow W = m \cdot \text{mass of solvent in kg} \]
\[ W = 0.667 \cdot 0.030 = 0.0200 \, \text{mol} \]

### Step 7: Calculate the number of moles of benzoic acid
Using the molar mass of benzoic acid:
\[ \text{Number of moles} = \frac{W}{\text{molar mass}} = \frac{W}{122} \]

### Step 8: Calculate the value of n
Since the degree of association is 80%, we can find the value of n:
\[ n = \frac{1}{1 - 0.8} = \frac{1}{0.2} = 5 \]

### Final Answer
The numerical value of n is 5.