1.22 g of an organic acid is separately dissolved in 100 g of benzene `(K_(b) = 2.6 K kg mol^(-1))` and 100 g of acetone `(K_(b) = 1.7 K kg mol^(-1))`. The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by `0.17^(@)C`. The increase in boiling point of solution in benzene in `""^(@)C` is `x xx 10^(-2)`. The value of x is ___________ . (Nearest integer)
[Atomic mass : `C = 12.0 , H = 1.0, O = 16.0`]

Answer

To solve the problem, we need to calculate the boiling point elevation for both benzene and acetone solutions of the organic acid. The acid dimerizes in benzene but remains as a monomer in acetone.

### Step-by-Step Solution:

1. **Determine the Molar Mass of the Organic Acid:**
Let the molar mass of the organic acid be \( M \) g/mol.

2. **Calculate the Molality in Acetone:**
The increase in boiling point in acetone is given as \( \Delta T_b = 0.17^\circ C \).
Using the formula for boiling point elevation:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where \( i \) is the van 't Hoff factor (which is 1 for monomers), \( K_b \) for acetone is 1.7 K kg/mol, and \( m \) is the molality.
Rearranging gives:
\[ m = \frac{\Delta T_b}{i \cdot K_b} = \frac{0.17}{1 \cdot 1.7} = 0.1 \, \text{mol/kg} \]

3. **Calculate the Number of Moles of the Acid in Acetone:**
The molality \( m \) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
With 100 g of acetone (0.1 kg), we can find the moles of the acid:
\[ \text{moles of acid} = m \cdot \text{kg of solvent} = 0.1 \cdot 0.1 = 0.01 \, \text{mol} \]

4. **Calculate the Molar Mass of the Acid:**
From the mass of the acid (1.22 g) and the number of moles (0.01 mol):
\[ M = \frac{\text{mass}}{\text{moles}} = \frac{1.22 \, \text{g}}{0.01 \, \text{mol}} = 122 \, \text{g/mol} \]

5. **Calculate the Molality in Benzene:**
In benzene, the acid dimerizes, so \( i = 0.5 \). Using the same boiling point elevation formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
We need to find the molality \( m \) in benzene:
\[ m = \frac{1.22 \, \text{g}}{122 \, \text{g/mol}} \cdot \frac{1000 \, \text{g}}{100 \, \text{g}} = \frac{0.01 \, \text{mol}}{0.1 \, \text{kg}} = 0.1 \, \text{mol/kg} \]

6. **Calculate the Boiling Point Elevation in Benzene:**
Now substituting the values into the boiling point elevation equation for benzene:
\[ \Delta T_b = 0.5 \cdot 2.6 \cdot 0.1 = 0.13^\circ C \]

7. **Express the Result:**
The increase in boiling point of the solution in benzene is \( 0.13^\circ C \), which can be expressed as:
\[ 0.13 = 13 \times 10^{-2} \]
Therefore, the value of \( x \) is \( 13 \).

### Final Answer:
The value of \( x \) is **13**.