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← JEE Advanced 2025 · Paper 2
ChemistryNumerical+4Question 45 of 48

JEE Advanced 2025 · Paper 2, Chemistry question 13

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\,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−32.00\ \mathrm{g\,dm^{-3}}, the molar mass of the macromolecule is calculated to be X×104 g mol−1\boldsymbol{X}\times10^4\ \mathrm{g\,mol^{-1}}. The value of X\boldsymbol{X} is _____.

Use: Universal gas constant (R) =8.3 J K−1 mol−1= 8.3\ \mathrm{J\,K^{-1}\,mol^{-1}} and acceleration due to gravity (gg) =10 m s−2= 10\ \mathrm{m\,s^{-2}}

Answer and worked solution

Official answer: 2.4

Osmotic pressure.
π=hρg=0.02×1000×10=200 Pa.\pi=h\rho g=0.02\times1000\times10=200\ \text{Pa}.

Molar mass. From π=cMRT\pi=\frac{c}{M}RT, with c=2.00c=2.00 g dm⁻³ =2000=2000 g m⁻³:
M=cRTπ=2000×8.3×300200=2.49×104 g mol−1.M=\frac{cRT}{\pi}=\frac{2000\times8.3\times300}{200}=2.49\times10^4\ \text{g mol}^{-1}.