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

2.05M solution of acetic acid means 2.05 moles per lit. of solutionweight of acetic acid = 2.05 x 60 = 123 g(no.of moles ×Molecular Wt.)Weight of solution = 1000 x 1.02 = 1020 g (volume×density)∴   Weight of water = (1020 - 123) = 897 g∴   molality=n×1000Wt. of solvent=2.05  × 1000897  =  2.285m(where , n= no. of moles of solute)