The density of a $2.0\mathrm{M}$ solution of acetic acid in water is $1.01\mathrm{g}{\mathrm{cm}}^{-3}$ . The molality of acetic acid in solution is 

For 2.0 solution, we will have 

Amount of acetic acid, $n_2=2.0\mathrm{mol}$     Volume of solution, $V=1000{\mathrm{cm}}^3$

Mass of solution  $\mathrm{m}=\mathrm{\rho V}=\left(1.01\mathrm{g}{\mathrm{cm}}^{-3}\right)\left(1000{\mathrm{cm}}^3\right)=1010\mathrm{g}$

Mass of acetic acid, ${\mathrm{m}}_2={\mathrm{n}}_2{\mathrm{M}}_2=(2.0\mathrm{mol})\left(60\mathrm{g}{\mathrm{mol}}^{-1}\right)=120\mathrm{g}$

Mass of water in $1010\mathrm{g}$ of solution ${\mathrm{m}}_1=1010\mathrm{g}-120\mathrm{g}=890\mathrm{g}=0.890\mathrm{kg}$ 

Molality of acetic acid  $=\frac{{\mathrm{n}}_2}{{\mathrm{m}}_1}=\frac{2.0\mathrm{mol}}{0.890\mathrm{kg}}=2.25\mathrm{molkg}$