Arrange the following aqueous solutions of urea in increasing order of their freezing point?
A. 2% (w/v)
B. 10% (w/v)
C. 8% (w/v)
D. 5% (w/v)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → B, C, D, A
Adding a non-volatile solute lowers the freezing point of a solvent. This depression in freezing point is a colligative property: it depends only on the number of solute particles, i.e. on the molality:
\( \displaystyle \Delta T_f=K_f\,m \)
The more concentrated the solution, the larger ΔTf and the lower its freezing point. Urea is a non-electrolyte (i = 1), and all four solutions contain the same solute in the same solvent, so the freezing point falls as the percentage rises.
Converting % (w/v) to molar concentration (molar mass of urea = 60 g mol−1):
\( \displaystyle 2\%:\ \frac{20\ \mathrm{g\,L^{-1}}}{60\ \mathrm{g\,mol^{-1}}}=0.33\ \mathrm{mol\,L^{-1}} \)
\( \displaystyle 5\%:\ \frac{50}{60}=0.83\ \mathrm{mol\,L^{-1}} \quad 8\%:\ \frac{80}{60}=1.33\ \mathrm{mol\,L^{-1}} \quad 10\%:\ \frac{100}{60}=1.67\ \mathrm{mol\,L^{-1}} \)
Depression in freezing point: 10% > 8% > 5% > 2%, so the freezing points increase in the order:
\( \displaystyle 10\% < 8\% < 5\% < 2\% \)
Matching the labels:
B. 10% (w/v): lowest freezing point
C. 8% (w/v)
D. 5% (w/v)
A. 2% (w/v): highest freezing point
Shortcut: The most concentrated solution B freezes lowest and the most dilute A freezes highest. Only Option (3) starts with B and ends with A.