Density of a 2.05 M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is:

Suppose the volume of solution is 1000 ml.

$1.02gm$ in 1 ml of soln.

$1020gm$ in 1000 ml of soln.

Number of moles $=2\cdot 05M\times 1lit$$=2.05mol$

Mass of solute $=n\times M\cdot wt$$=2.05\times 60=123gm$

Mass of solvent $=1020-123=897gm$

Molality $=\frac{123}{60}\times \frac{1000}{897}$$=2.28molk{g}^{-1}$