Density of 2.05M solution of acetic acid in water is 1.02 g/mL, the molality of that solution is
 

let us assume the volume of solution to be 1000ml
as given in the question 1.02gm in 1 ml of solution.
1020gm in 1000ml of solution.
number of moles =$2.05\mathrm{M}\times 1\mathrm{lit}=2.05 \mathrm{mol}$
mass of solute =$\mathrm{n}\times \mathrm{Mol}.\mathrm{wt}.=123\mathrm{gm}$

mass of solvent=1020-123=897gm

molarity=(123/60)*(1000/891)=2.28mol/kg~2.3mol/kg.

Hence, option(A) 2.3 is correct.