Molar mass of a solute?

For benzene :
Freezing point = 5.5°C
Cryoscopic constant, Kf = 5.12 °C•kg/mol

Acetic acid dissolves in benzene in molecular form, and thus van’t Hoff factor, i = 1
Hence, ΔTf = Kf • m
Hence, molality of the solute, m = ΔTf / Kf = [(5.5 - 3.5)°C] / (5.12 °C•kg/mol) = 0.391 mol/kg

No. of moles of solute in the solution = (0.391 mol/kg) × (80/1000 kg) = 0.0313 mol
Molar mass of the solute = (3.8 g) / (0.0313 mol) = 120 g/mol (to 2 sig. fig.)


(2 sig. fig. is taken due to the lowest sig. fig. in 3.8 g is 2.)
(The molar mass of the solute, acetic acid, is 60 g/mol, only a half of the calculated value.)
(It is believed acetic acid exists as dimer, i.e. 2 molecules linked together by hydrogen bonds, in the solution.)
(To calculate the exact molar mass of acetic acid, the van’t Hoff factor should be amended to 0.5.)
(The structure of the dimer of acetic acid is shown below.)

(5.49 - 3.5)°C / (5.12 °C/m) = 0.38867 m

(0.38867 mol/kg) x (0.080 kg) = 0.03109 mol

(3.8 g) / (0.03109 mol) = 122 g/mol

This is about twice the actual molecular mass of acetic acid (60.0 g/mol) so it appears that acetic acid forms a dimer in benzene. Or saying it another way, in benzene, acetic acid has a van't Hoff factor near 0.5