Q13 / 48 · JEE Advanced 2025 (Paper 2) · Chemistry
At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height ( h ) of the solution (density $=1.00 \mathrm{~g} \mathrm{~cm}^{-3}$ ) where h is equal to 2.00 cm . If the concentration of the dilute solution of the macromolecule is $2.00 \mathrm{~g} \mathrm{dm}^{-3}$, the molar mass of the macromolecule is calculated to be $\boldsymbol{X} \times 10^4 \mathrm{~g} \mathrm{~mol}^{-1}$. The value of $\boldsymbol{X}$ is $\qquad$ .
Use : Universal gas constant $(\mathrm{R})=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ and acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} \mathrm{~s}^{-2}$
Correct answer: 2.49
Solution
$\begin{aligned} & \pi=\rho g h=10^3 \times 10 \times 2 \times 10^{-2} \text { Pascal }=200 \text { Pascal } \\ & \pi=C R T \\ & 200=\frac{2}{M} \times 1000 \times 8.3 \times 300 \\ & M=24900=2.49 \times 10^4 \mathrm{~g} / \mathrm{mol} \\ & X=2.49\end{aligned}$
Hence, the correct answer is 2.49.