In benzene, ethanoic acid molecules can join in pairs…

Step 1. Molality. $m = \dfrac{1.20/60}{0.0300} = 0.667\ \text{mol/kg}$.

Step 2. The fall for single molecules. $5.12\ \text{K kg/mol} \times 0.667\ \text{mol/kg} = 3.41\ \text{K}$.

Step 3. The factor. $i = \dfrac{1.88\ \text{K}}{3.41\ \text{K}} = 0.551$. It is below $1$, so the acid joins.

Step 4. The share in pairs. $\alpha = 2(1 - 0.551) = 0.898$.