$3.0 g$ of ethanoic acid (molar mass $60 g mo l^{- 1}$ ) is dissolved in $100 g$ of benzene, where it partly dimerises: $2 HA ⇌ (HA)_{2}$ . The observed depression of freezing point is $1.50 K$ ( $K_{f} = 5.12 K kg mo l^{- 1}$ ). Define the equilibrium constant in terms of molalities, $K_{m} = m_{dimer} / m_{monomer}^{2}$ . Find $K_{m}$ in $kg mo l^{- 1}$ , to one decimal place.

Numerical answer type, answer in kg/mol, tolerance 0.2.

Answer: 28 kg/mol

Formal molality

m_{0} = (3.0/60) /0.100 = 0.500

. Particle molality

= 1.50/5.12 = 0.29297

, so

i = 0.5859 = 1 - α /2

and

α = 0.8281

. Then

m_{monomer} = m_{0} (1 - α) = 0.08594

and

m_{dimer} = m_{0} α /2 = 0.20703

K_{m} = 0.20703/ (0.08594)^{2} = 28.033 \approx 28.0 kg mo l^{- 1}