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JEE Advanced · Chemistry · 16. 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 \(=1.00 \mathrm{~g} \mathrm{~cm}^{-3}\) ) 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 \(\qquad\) .
Use : Universal gas constant \((\mathrm{R})=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) and acceleration due to gravity \((\mathrm{g})=10 \mathrm{~m} \mathrm{~s}^{-2}\)

  1. A 2.14
  2. B 2.19
  3. C 2.49
  4. D 5.45
Verified Solution

Answer & Solution

Correct Answer

(C) 2.49

Step-by-step Solution

Detailed explanation

\(\pi=\rho g h=10^3 \times 10 \times 2 \times 10^{-2}\) \(\text {Pascal}=200 \text { Pascal } \)
\( \pi=C R T \)
\( 200=\frac{2}{M} \times 1000 \times 8.3 \times 300 \)
\( M=24900=2.49 \times 10^4 \mathrm{~g} / \mathrm{mol} \)
\( X=2.49\)
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