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:
(1) $P_{2}>P_{1}>P_{3}$
(2) $\mathrm{P}_{1}>\mathrm{P}_{2}>\mathrm{P}_{3}$
(3) $\mathrm{P}_{2}>\mathrm{P}_{3}>\mathrm{P}_{1}$
(4) $\mathrm{P}_{3}>\mathrm{P}_{1}>\mathrm{P}_{2}$

contexto.130

3 years ago

Answer is (1)

Osmotic pressure is given as $\pi=i C R T$

where $C$ is molar concentration of the solution

With increase in molar concentration of solution osmotic pressure increases.

Because the weight of all solutes and their solution volume are identical, the higher the molar mass of the solute, the lower the molar concentration, and the lower the osmotic pressure.

Order of molar mass of solute decreases as Sucrose $>$ Glucose $>$ Urea

So, correct order of osmotic pressure of solution is $P_{3}<P_{1}<P_{2}$