The molality of a urea solution in which 0.0200 g of urea $\left({\mathrm{NH}}_2{\mathrm{CONH}}_2\right)$ is added to 0.400 $d{m}^3$ of water at STP is 

$\begin{array}{l}1\mathrm{\hspace{0.17em}}{\mathrm{dm}}^3=1000\mathrm{\hspace{0.17em}}{\mathrm{cm}}^3=1\mathrm{L}=1 \mathrm{kg}\\ 0.400\hspace{0.17em}{\mathrm{dm}}^3 \mathrm{of} {\mathrm{H}}_2\mathrm{O} \mathrm{at} \mathrm{STP} = 0.400 \mathrm{kg} \mathrm{of} {\mathrm{H}}_2\mathrm{O}\mathrm{\hspace{0.17em}}(\mathrm{density} = 1.0 \mathrm{g} /\mathrm{cm})\\ \mathrm{Moles} \mathrm{of} \mathrm{urea} =\frac{0.0200}{60}\\ \therefore \hspace{0.17em}\mathrm{Molality}=\frac{\mathrm{Moles} \mathrm{of} \mathrm{solute} (\mathrm{urea})}{\mathrm{kg} \mathrm{of} \mathrm{solvent}}\\ =\frac{0.020/60}{0.400}\\ =8.33\times {10}^{-4} \mathrm{mol} {\mathrm{kg}}^{-1}\end{array}$