Density of 2.05M solution of acetic acid in water is 1.02 g/mL, the molality of that solution is ___. (one decimal)

Assume the solution has a 1000 ml volume.

Soln contains 1.02g in 1ml.
1000ml of soln with 1020g.
amount of moles $=2·05\mathrm{M}\times 1lit=2.05 \mathrm{mol}$
Mass of solute $=n\times M·wt=2.05\times 60=123\mathrm{gm}$
Mass of solvent =1020-123=897 gm
Molality $=\frac{123}{60}\times \frac{1000}{897}=2.28 \mathrm{mol}{ \mathrm{kg}}^{-1}$.

Hence, the required answer is 2.3.