JEE Advanced 2025 Paper 2 · Q13 · Solutions & Colligative Properties

Question

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 $g/cm^3$) where $h$ is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 $g/dm^3$, the molar mass of the macromolecule is calculated to be $X \times 10^4$ g/mol. The value of X is __________.

Use: Universal gas constant $R = 8.3$ $J K^{-1} mol^{-1}$ and acceleration due to gravity $g = 10$ $m/s^2$.

SolveKar AI · Accepted Answer

Osmotic pressure $\pi = h \cdot ho \cdot g = 0.02 	ext{ m} \cdot 1000 	ext{ kg/}m^3 \cdot 10 	ext{ m/}s^2 = 200$ Pa. 
van 't Hoff: $\pi = (c/M) R T 	o M = c R T/\pi$. 
$c = 2$ $g/dm^3 = 2$ kg/$m^3$ (since 1 g/L = 1 kg/$m^3$). 
$M = (2 	imes 8.3 	imes 300)/200 = 4980/200 = 24.9$ kg/mol = $2.49 	imes 10^4$ g/mol. 
Hence $X = 2.49$.