12.2 g of benzoic acid (molar mass 122 g mol^-1) is dissolved in 250 g of benzene (Kf=5.12 K kg mol^-1). Assuming the acid does not ionise in benzene…

Molality of the formal solute $m=(12.2/122)/0.250=0.400\ \mathrm{mol\,kg^{-1}}$. From $\Delta T_f=iK_fm$, $i=1.20/(5.12\times0.400)=0.5859$. For association $2\mathrm{A}\rightleftharpoons\mathrm{A_2}$, starting from $1$ mole of monomer units, if a fraction $\alpha$ is dimerised the total number of moles of particles is $1-\alpha+\alpha/2=1-\alpha/2$, so $i=1-\alpha/2$. Therefore $\alpha=2(1-i)=2(1-0.5859)=0.8281$, i.e. $82.81\%$ of the molecules exist in dimer form. The result lies between $0$ and $1$, which confirms that the assumption of dimerisation alone is consistent. Check on the data: with $i=0.586$ the calculated $\Delta T_f$ for no association would be $5.12\times0.4=2.05\ \mathrm{K}$, and the observed $1.20\ \mathrm{K}$ is lower, as expected for dimerisation, which reduces the number of particles.