610 g of benzoic acid (molar mass 122 g mol^-1) is dissolved in 30.0 g of benzene (cryoscopic constant Kf=5.12 K kg mol^-1). The solute partly…

First find the stoichiometric molality: $n=0.610/122=5.00\times10^{-3}$ mol in $0.0300$ kg, so $m=0.1667$ molal. If nothing associated, $\Delta T_f=5.12\times0.1667=0.853$ K. The van't Hoff factor is $i=0.512/0.853=0.600$. For dimerisation with degree of association $\alpha$, the total particles are $1-\alpha/2$ per formula unit, so $1-\alpha/2=0.600$ and $\alpha=0.80$. Equilibrium molalities: $m(\mathrm{M})=0.1667(1-0.8)=0.03333$ and $m(\mathrm{M_2})=0.1667\times0.8/2=0.06667$. Then $K_m=0.06667/(0.03333)^2=60\ \mathrm{kg\,mol^{-1}}$. The value 120 arises from counting $m(\mathrm{M_2})$ as $0.1333$ (no factor of one half for the dimer), while 30 and 15 come from mis-assigning the monomer fraction after association, so always write the equilibrium amounts explicitly before forming the ratio.