0 g of acetic acid (M=60 g mol^-1) is dissolved in 200 g of benzene (Kf=5.12 K kg mol^-1). The acid associates to dimers in benzene, and 80% of the…

Moles of acetic acid $=6.0/60=0.10$. Molality $m=0.10/0.200=0.50\ \mathrm{mol\,kg^{-1}}$. For association $2\mathrm{CH_3COOH}\rightleftharpoons(\mathrm{CH_3COOH})_2$ with degree of association $\alpha=0.80$, the van't Hoff factor is $i=1-\alpha/2=1-0.40=0.60$, because each 100 particles initially give $100-80+40=60$ particles. Then $\Delta T_f=i\,K_f\,m=0.60\times5.12\times0.50=1.536\ \mathrm{K}\approx1.54\ \mathrm{K}$. Note that the factor is less than one (fewer particles than formal molecules), so a naive calculation with $i=1$ gives $2.56$ K, which is too large. A common error is to use $i=1+\alpha/2$ or $i=2$, as if the solute dissociated; here association halves the number of particles, and the correct factor is $0.60$. The answer is $1.54$ K.