Previous Year NEET Question- Solution L-1

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 solute
• With increase in molar concentration of solution osmotic pressure increases.
• Since the weight of all solutes and the solution volume are equal, a higher molar mass of solute will result in a smaller molar concentration and a smaller osmotic pressure.
• Order of molar mass of solute increases as Sucrose > Glucose > Urea
• So, correct order of osmotic pressure of solution is $\mathrm{P}_3>\mathrm{P}_1>\mathrm{P}_2$