Question

Estimate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.

Answer 1

Explanation:

Mass of solution given is $2.5\mathrm{Kg}$ and

Number of moles of urea $0.25\mathrm{m}$

The molar mass of urea$\left(N{H}_{2}CON{H}_2\right)$

$$\begin{gathered} {\mathrm{NH}}_2{\mathrm{CONH}}_2=(14+2\times 1+12+16+14+2\times 1) \\ =60\mathrm{g}{\mathrm{mol}}^{-1} \end{gathered}$$

Therefore the mass of urea will be $\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{urea}\times \mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{urea}$

And, the mass of $0.25$moles of urea will be,

$$\begin{gathered} 0.25\mathrm{mol}\times 60\mathrm{g}{\mathrm{mol}}^{-1} \\ =15\mathrm{g} \end{gathered}$$

Thus, the mass of the solution is$1000\mathrm{g}+15\mathrm{g}=1015\mathrm{g}$

$$\begin{gathered} 1015\mathrm{g}\mathrm{of}\mathrm{aqueous}\mathrm{solution}\mathrm{contains}\mathrm{urea}=15\mathrm{g} \\ 2500\mathrm{g}\mathrm{of}\mathrm{aqueous}\mathrm{will}\mathrm{requires}\mathrm{urea}=\frac{15}{1015\mathrm{kg}\times 2500\mathrm{g}}=36.95\mathrm{g} \end{gathered}$$

Therefore, the mass of urea is required in making $2.5\mathrm{Kg}\mathrm{of}0.25\mathrm{molal}\mathrm{aqueous}\mathrm{solution}\mathrm{is}36.95\mathrm{g}$.