JEE Advanced 2023 Paper 2 · Q12 · Solutions & Colligative Properties

Question

$50$ mL of $0.2$ molal urea solution (density $= 1.012$ g·mL$^{-1}$ at $300$ K) is mixed with $250$ mL of a solution containing $0.06$ g of urea. Both solutions were prepared in the same solvent. The osmotic pressure (in Torr) of the resulting solution at $300$ K is ___. [Molar mass urea $= 60$ g·mol$^{-1}$; $R = 62$ L·Torr·K$^{-1}$·mol$^{-1}$. Assume $\Delta H_{	ext{mix}} = \Delta V_{	ext{mix}} = 0$]

SolveKar AI · Accepted Answer

Solution 1 mass = $50\cdot 1.012 = 50.6$ g. From $0.2$ molal: per kg solvent has $0.2$ mol urea. Let $x$ = mass urea, $50.6 - x$ = mass solvent. Then $x/60 \cdot 1000/(50.6-x) = 0.2 \Rightarrow x \approx 0.6$ g (about $0.01$ mol urea). Solution 2: $0.06$ g $= 0.001$ mol urea. Total moles urea = $0.011$. Total volume (assuming additive) ≈ $50 + 250 = 300$ mL = $0.3$ L. Osmotic pressure $\pi = (n/V)RT = (0.011/0.3)(62)(300) = (0.011/0.3)\cdot 18600 = 682$ Torr.