`12.2 g` of benzoic acid `(Mw=122)` in `100 g` water has elevation in boiling point of `0.27.K_(b)=0.54 K kg mol^(-1)`.If there is `100%` polymerization, the number of molecules of benzoic acid in associated state is

Options

A. 2
B. 1
C. 3
D. 4

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of benzoic acid
Given:
- Mass of benzoic acid = 12.2 g
- Molar mass of benzoic acid (Mw) = 122 g/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{12.2 \, \text{g}}{122 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 2: Calculate the mass of the solvent in kg
Given:
- Mass of water = 100 g

Convert grams to kilograms:
\[ \text{Mass of water} = \frac{100 \, \text{g}}{1000} = 0.1 \, \text{kg} \]

### Step 3: Calculate the molality (m)
Molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \, \text{mol}}{0.1 \, \text{kg}} = 1 \, \text{mol/kg} \]

### Step 4: Use the elevation in boiling point formula
The elevation in boiling point formula is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \(\Delta T_b = 0.27 \, \text{K}\)
- \(K_b = 0.54 \, \text{K kg mol}^{-1}\)
- \(m = 1 \, \text{mol/kg}\)

Substituting the known values:
\[ 0.27 = i \cdot 0.54 \cdot 1 \]
Solving for \(i\):
\[ i = \frac{0.27}{0.54} = 0.5 \]

### Step 5: Relate \(i\) to the degree of association
Using the formula for the van 't Hoff factor \(i\) in the case of association:
\[ i = 1 + \frac{1}{n - 1} \cdot \alpha \]
Where \(\alpha = 1\) (100% polymerization).

Substituting the values:
\[ 0.5 = 1 + \frac{1}{n - 1} \cdot 1 \]
Rearranging gives:
\[ 0.5 - 1 = \frac{1}{n - 1} \]
\[ -0.5 = \frac{1}{n - 1} \]
Taking the reciprocal:
\[ n - 1 = -2 \implies n = 2 \]

### Conclusion
The number of molecules of benzoic acid in the associated state is \(n = 2\).

### Final Answer
The answer is **Option A: 2**.

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