1. Use the acetone data to find molar mass of the acid

Since the acid remains monomeric in acetone, the van't Hoff factor is $i=1$.

Boiling point elevation formula:

$\Delta T_b = i K_b m$

Given:

• $\Delta T_b = 0.17^\circ C$
• $K_b = 1.7\ \text{K kg mol}^{-1}$
• mass of solvent $=100\,g = 0.1\,kg$
• mass of solute $=1.22\,g$

So,

$m = \frac{\Delta T_b}{K_b} = \frac{0.17}{1.7} = 0.1\ \text{mol kg}^{-1}$

Now moles of acid in $0.1\,kg$ acetone:

$n = m \times 0.1 = 0.1 \times 0.1 = 0.01\ \text{mol}$

Hence molar mass of acid:

$M = \frac{1.22}{0.01} = 122\ \text{g mol}^{-1}$

2. Find molality in benzene if no association occurred

Same mass of acid is dissolved in $100\,g = 0.1\,kg$ benzene.

Moles of acid are again:

$n = \frac{1.22}{122} = 0.01\ \text{mol}$

Thus actual analytical molality is

$m = \frac{0.01}{0.1} = 0.1\ \text{mol kg}^{-1}$

3. Account for dimerization in benzene

The acid dimerizes in benzene. Assuming complete dimerization into dimers, number of solute particles becomes half.

Therefore van't Hoff factor:

$i = \frac{1}{2}$

So,

$\Delta T_b = iK_bm = \frac{1}{2}\times 2.6 \times 0.1$

$\Delta T_b = 0.13^\circ C$

4. Match with the required form

Given:

$\Delta T_b = x \times 10^{-2}\ ^\circ C$

Since

$0.13 = 13 \times 10^{-2}$

we get

$x=13$

5. Comparison with stored answer

Derived answer: $13$

Stored correct answer: $13$

They agree.