JEE Advanced 2021 Paper 1 Chemistry: Paragraph 1

If no association took place, $m_0=\dfrac{6.10/122}{0.100}=0.500\ \mathrm{mol\,kg^{-1}}$ and $\Delta T_f=K_fm_0=2.56\ \mathrm{K}$. The van't Hoff factor is therefore $i=\dfrac{1.536}{2.56}=0.60$. For association into dimers, if a fraction $\alpha$ of the monomers dimerises, the number of particles per initial monomer is $1-\alpha+\alpha/2=1-\alpha/2$. Thus $1-\alpha/2=0.60$, giving $\alpha=0.80$. Association lowers the number of solute particles, which is why $i<1$ and the observed depression is smaller than the calculated one. A negative or greater-than-unity value of $\alpha$ would signal an arithmetic slip, since the fraction of associated molecules must lie between zero and one, and the answer is checked by recomputing $0.5(1-0.4)\times5.12=1.536\ \mathrm{K}$.

From the first part, $\alpha=0.80$ and $m_0=0.500\ \mathrm{mol\,kg^{-1}}$. At equilibrium the molality of free monomer is $m_0(1-\alpha)=0.500\times0.20=0.100\ \mathrm{mol\,kg^{-1}}$. Every two monomer units that associate produce one dimer, so the dimer molality is $m_0\alpha/2=0.500\times0.80/2=0.200\ \mathrm{mol\,kg^{-1}}$. Then $K_m=\dfrac{0.200}{(0.100)^2}=20\ \mathrm{kg\,mol^{-1}}$. A common error is to use $m_0\alpha$ for the dimer, which ignores the 2:1 stoichiometry and gives 40. The total particle molality $0.100+0.200=0.300$ matches $i\,m_0=0.60\times0.500$, which confirms the calculation is self-consistent. The large value of $K_m$ shows that the dimer is strongly favoured in a non-polar solvent such as benzene, where hydrogen-bonded carboxylic acid pairs are stable and there is little competing solvation of the monomer.