Question: The following solutions were prepared by dissolving $10 \mathrm{~g}$ of glucose $\left(\mathrm{C}6 \mathrm{H}{12} \mathrm{O}_6\right)$ in $250 \mathrm{ml}$ of water $\left(\mathrm{P}_1\right), 10 \mathrm{~g}$ of urea $\left(\mathrm{CH}4 \mathrm{~N}2 \mathrm{O}\right)$ in $250 \mathrm{ml}$ of water $\left(\mathrm{P}2\right)$ and $10 \mathrm{~g}$ of sucrose $\left(\mathrm{C}{12} \mathrm{H}{22} \mathrm{O}{11}\right)$ in $250 \mathrm{ml}$ of water $\left(\mathrm{P}_3\right)$. The right option for the decreasing order of osmotic pressure of these solutions is :
A) $P_3>P_1>P_2$
B) $P_2>P_1>P_3$
C) $P_1>P_2>P_3$
D) $P_2>P_3>P_1$
Answer: $P_2>P_1>P_3$
Solution:
- Osmotic pressure $(\pi)=$ iCRT where $\mathrm{C}$ is molar concentration of the solution
- With increase in molar concentration of solution osmotic pressure increases.
- Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.
- Order of molar mass of solute decreases as Sucrose > Glucose > Urea
- So, correct order of osmotic pressure of solution is $\mathrm{P}_3>\mathrm{P}_1>\mathrm{P}_2$
