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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Chemistry
Chapter
Solutions

JEE Advanced 2025 · Paper 2 · Chemistry · Solutions

At 300 K , an ideal dilute solution of a macromolecule exerts...

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=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 gdm−32.00 \mathrm{~g} \mathrm{dm}^{-3}, the molar mass of the macromolecule is calculated to be X×104 g mol−1\boldsymbol{X} \times 10^4 \mathrm{~g} \mathrm{~mol}^{-1}. The value of X\boldsymbol{X} is \qquad . Use : Universal gas constant (R)=8.3 J K−1 mol−1(\mathrm{R})=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} and acceleration due to gravity (g)=10 m s−2(\mathrm{g})=10 \mathrm{~m} \mathrm{~s}^{-2}

Correct answer

0

Step-by-step solution

π=ρgh=103×10×2×10−2 Pascal =200 Pascal π=CRT200=2M×1000×8.3×300M=24900=2.49×104g/molX=2.49\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}
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