JEE Advance - Chemistry (2025 - Paper 2 Online - No. 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 $=1.00 \mathrm{~g} \mathrm{~cm}^{-3}$ ) 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 $(R)=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ and acceleration due to gravity $(g)=10 \mathrm{~m} \mathrm{~s}^{-2}$
Explanation
Density of the solution:
$ \text{Density} = 1.00 \, \text{g cm}^{-3} = 1000 \, \text{kg m}^{-3} $
Height of the solution column ($h$):
$ h = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} $
Acceleration due to gravity ($g$):
$ g = 10 \, \text{m s}^{-2} $
Calculating osmotic pressure ($\pi$):
Osmotic pressure is given by the equation:
$ \pi = h \cdot \rho \cdot g $
Substituting the known values:
$ \pi = 2 \times 10^{-2} \times 1000 \times 10 = 200 \, \text{N m}^{-2} $
Relating osmotic pressure to concentration and molar mass:
The equation for osmotic pressure in terms of concentration and molar mass is:
$ \pi = cRT $
where $c = \frac{2000}{M}$ as the concentration in g/dm$^3$ needs to be converted to mol/dm$^3$ by dividing by the molar mass $M$. Incorporating the given temperature and universal gas constant, we have:
$ 200 = \left(\frac{2000}{M}\right) \cdot 8.3 \cdot 300 $
Solving for the molar mass $M$:
Rearrange and solve the equation to find $M$:
$ M = 24900 \, \text{g mol}^{-1} = 2.49 \times 10^4 \, \text{g mol}^{-1} $
Thus, the value of $X$ is $2.49$.
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