What mass of urea (NH2CONH2) must be dissolved in 500 g of water to lower its freezing point by 0.186 °C? (Kf for water = 1.86 K kg mol−1; molar mass of urea = 60 g mol−1)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3 g
For a dilute solution of a non-electrolyte such as urea (i = 1), the depression of freezing point is proportional to the molality m of the solute:
\( \displaystyle \Delta T_f=K_f\,m \)
\( \displaystyle m=\frac{\Delta T_f}{K_f}=\frac{0.186\,\mathrm{K}}{1.86\,\mathrm{K\,kg\,mol^{-1}}}=0.1\,\mathrm{mol\,kg^{-1}} \)
Moles of urea needed in 500 g (0.5 kg) of water, and their mass:
\( \displaystyle n=0.1\,\mathrm{mol\,kg^{-1}}\times0.5\,\mathrm{kg}=0.05\,\mathrm{mol} \)
\( \displaystyle w=0.05\,\mathrm{mol}\times60\,\mathrm{g\,mol^{-1}}=3\,\mathrm{g} \)