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

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Solutions

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}

Answer: 2.49

Step 1: Calculate osmotic pressure from height:

π = ρ g h = 1 0^{3} \times 10 \times 2 \times 1 0^{- 2} = 200 Pa

Step 2: Use the osmotic pressure equation:

π = C R T

where C is molar concentration = (mass concentration)/M

π = \frac{w}{M} \times 1000 \times R T

Step 3: Solve for molar mass:

200 = \frac{2}{M} \times 1000 \times 8.3 \times 300

M = \frac{2 \times 1000 \times 8.3 \times 300}{200} = \frac{4 , 980 , 000}{200} = 24, 900 g/mol

M = 2.49 \times 1 0^{4} g/mol

Therefore, X = 2.49

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer