22 g of benzoic acid (M=122 g mol^-1) is dissolved in 50.0 g of benzene (Kf=5.12 K kg mol^-1). Assuming the acid exists only as monomers and dimers…

Molality assuming no association: $m=(1.22/122)/0.0500=0.200\ \mathrm{mol\,kg^{-1}}$, so the expected depression is $5.12\times0.200=1.024$ K. Observed is $0.64$ K, hence $i=0.64/1.024=0.625$ (A is correct). For dimerisation, if a fraction $\alpha$ of the monomers associates, total particles $=1-\alpha/2$ per mole of monomer, so $\alpha=2(1-i)=0.75$, i.e. $75\%$, not $62.5\%$ (B is wrong; it confuses $i$ with $\alpha$). Association reduces the number of particles, so the apparent molar mass is $122/0.625=195.2\ \mathrm{g\,mol^{-1}}$, not $76.3$ (C is wrong). Species per mole of monomer units: monomer $0.25$, dimer $0.375$, total $0.625$, so the dimer fraction is $0.375/0.625=0.60$ (D is wrong). Only A is correct.