The following solutions were prepared by dissolving 10 g glucose ($\mathrm{C}{}_6\mathrm{H}{}_{12}\mathrm{O}{}_6$) in 250 mL of water (P1), 10 g of urea ($\mathrm{CH}{}_4\mathrm{N}{}_2\mathrm{O}$) in 250 mL of water (P2) and 10 g of sucrose ($\mathrm{C}{}_{12}\mathrm{H}{}_{22}\mathrm{O}{}_{11}$) in 250 mL of water (P3). The right option for the decreasing order of osmotic pressure of these solutions is:

Solution & Explanation

\textbf{Answer:} (C) P2 > P1 > P3 \textbf{Solution:} Osmotic pressure is given by $\pi =\frac{wRT}{MV}$. Here the mass ($w=10$ g) and the volume ($V=250$ mL) are the same for all three, and all are non-electrolytes ($i=1$). Therefore $\pi \propto \frac{1}{M}$ — the osmotic pressure is inversely proportional to the molar mass. Molar masses: urea (P2) $=60$ g/mol $<$ glucose (P1) $=180$ g/mol $<$ sucrose (P3) $=342$ g/mol. Smallest molar mass gives the largest osmotic pressure, so the decreasing order is P2 > P1 > P3.