JEE Advanced 2025 — Paper 2 — CHEM-III solution

JEE Advanced2025Paper 2CHEM-III

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 $X \times 1 0^{4}$ g mol $^{- 1}$ . 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 $^{- 2}$

Your answer

Solution

Osmotic pressure from a liquid column: $π = ρ g h$ with $ρ = 1000$ kg m $^{- 3}$ , $g = 10$ m s $^{- 2}$ , $h = 2.00$ cm $= 2 \times 1 0^{- 2}$ m:

$π = 1000 \times 10 \times 2 \times 1 0^{- 2} = 200$ Pa.

Van't Hoff equation: $π = \frac{c}{M} R T$ where $c = 2.00$ g dm $^{- 3} = 2.00$ kg m $^{- 3}$ , $T = 300$ K:

$200 = \frac{2.00}{M} \times 8.3 \times 300$

$M = \frac{2.00 \times 8.3 \times 300}{200} = \frac{4980}{200} = 24.9$ kg mol $^{- 1} = 24900$ g mol $^{- 1}$

$M = 2.49 \times 1 0^{4}$ g mol $^{- 1}$ , so $X = 2.49$ .

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