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 $$x \times 10^{-2}$$. The value of x is _________. (Nearest integer)
[Atomic mass : C = 12.0, H = 1.0, O = 16.0]

We have an organic acid of mass $$1.22\ \text{g}$$ which is dissolved in each case in $$100\ \text{g}=0.1\ \text{kg}$$ of solvent. For any solution the elevation in boiling point is given by the formula

$$\Delta T_b = i\,K_b\,m$$

where $$\Delta T_b$$ = rise in boiling point, $$i$$ = van’t Hoff factor, $$K_b$$ = molal elevation constant of the solvent, $$m$$ = molality of the solution.

First we use the data for acetone, because in acetone the acid remains as a monomer (no association), so $$i=1$$.

The observed elevation is $$\Delta T_b = 0.17^{\circ}\text{C}$$ and for acetone $$K_b = 1.7\ \text{K kg mol}^{-1}$$.

The molality is

$$m=\frac{\text{moles of solute}}{\text{kg of solvent}} =\frac{w/M}{0.1} =\frac{1.22/M}{0.1} =\frac{12.2}{M}\ \text{mol kg}^{-1}$$

Substituting in the formula

$$0.17 = (1)\,(1.7)\left(\frac{12.2}{M}\right)$$ $$\Rightarrow 0.17 = \frac{20.74}{M}$$ $$\Rightarrow M = \frac{20.74}{0.17}$$ $$\Rightarrow M \approx 122\ \text{g mol}^{-1}$$

So the molar mass of the acid is $$122\ \text{g mol}^{-1}$$.

Now we calculate the molality explicitly for later use:

$$\text{Moles of acid} = \frac{1.22}{122}=0.01\ \text{mol}$$ $$m = \frac{0.01}{0.1}=0.1\ \text{mol kg}^{-1}$$

Next we deal with the benzene solution. In benzene the acid is known to dimerise:

$$2\,\text{A}\ \rightleftharpoons\ (\text{A})_2$$

For association of $$n$$ identical molecules, if the degree of association is $$\alpha$$, the van’t Hoff factor is

$$i = 1-\alpha+\frac{\alpha}{n}$$

Here $$n=2$$ (dimer) and, in non-polar benzene, the association is effectively complete, so $$\alpha \approx 1$$.

Hence

$$i = 1-\frac{1}{2}=0.5$$

Using $$K_b(\text{benzene}) = 2.6\ \text{K kg mol}^{-1}$$ and the previously found molality $$m=0.1$$, we get

$$\Delta T_b = i\,K_b\,m = (0.5)\,(2.6)\,(0.1) = 0.13^{\circ}\text{C}$$

The question states the elevation in benzene as $$x \times 10^{-2}\,^{\circ}\text{C}$$. Writing $$0.13^{\circ}\text{C}$$ in this form,

$$0.13^{\circ}\text{C}=13\times10^{-2}\,^{\circ}\text{C}$$

Thus $$x = 13$$ (already the nearest integer).

So, the answer is $$13$$.