2.05M solution of acetic acid in water means in 1 litre solution 2.05 moles of CH₃COOH present.

Density of solution = 1.02 g/ml (Given)

Assume the volume of solution = 1 litre = 1000 ml

$$\therefore$$ Mass of solution = 1000 $$ \times $$ 1.02 = 1020 gm

Molar mass of CH₃COOH = 60

So Mass of CH₃COOH (m_solute) = 2.05 $$ \times $$ 60 = 123

$$\therefore$$ Mass of solvent (m_solvent) = 1020 - 123 = 897 gm = 0.897 kg

Formula of molality (m) = $${{no\,of\,moles\,of\,solute} \over {weight\,of\,solvent\,in\,kg}}$$

$$\therefore$$ m = $${{2.05} \over {0.897}}$$ = 2.28

Using Formula :

Molality (m) = $${{1000 \times M} \over {1000 \times d - M \times {M_{solute}}}}$$

Here M = molarity, M_solute = molecular mass of solute, d = density of solution

$$\therefore$$ m = $${{1000 \times 2.05} \over {1000 \times 1.02 - 2.05 \times 60}}$$ = 2.28