Question

The following solutions were prepared by dissolving 10g of glucose $\left(C_6{H}_{12}{O}_6\right)$ in 250ml of water $\left(P_1\right)$ 10g of urea $\left(C{H}_4{N}_{2}O\right)$ in 250ml of water $\left(P_2\right)$ and 10g of sucrose $\left(C_{12}{H}_{12}{O}_{11}\right)$ in 250ml of water $\left(P_3\right)$. The right option for the decreasing order of osmotic pressure of these solutions is:

• A. $P_3>P_1>P_2$
• B. $P_2>P_3>P_1$
• C. $P_1>P_2>P_3$
• D. $P_2>P_1>P_3$

Answer 1

The correct option is D $P_2>P_1>P_3$

Osmotic pressure: $\left(\pi \right)$
$\pi =C\times R\times T$
Here
$C\text{is concentration (Molarity)}{\text{(mol L}}^{-1}\left)R\text{is ideal gas constant}{\text{(L atm mol}}^{-1}{\text{K}}^{-1}\right)T\text{is temperature (K)}$

$\pi =\frac{w}{M\times V}\times R\times T$
Since volume of solvent and weight of solute is same
$\pi \propto \frac{1}{\text{Molar mass}}$
Molar mass of Glucose is 180
Molar mass of Urea is 60
Molar mass of Sucrose is 342
${\pi }_{\text{Urea}}>{\pi }_{\text{Glucose}}>{\pi }_{\text{Sucrose}}{P}_2>P_1>P_3$
Option (a) is correct