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Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Liquid Solutions
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$200 cm^3$ of an aqueous solution of a protein contains 1.26g of the protein. The osmotic pressure of this solution at 300 K is found to be $2.57 \times 10^{−3}$ bar. Calculate the molar mass of the protein.

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Here, we are given: Mass of the solute (protein), $W_2 = 1.26g$
Volume of the solution (V) = $200 cm^3$ = 0.200 L
$\pi=2.57\times 10^{-3}$bar,T=300K,$R=0.083Lbar K^{-1}mol^{-1}$
Substituting these values in the formula $M_2=\large\frac{W_2RT}{\pi V}$
We get
$M_2=\large\frac{1.26g\times 0.083LbarK6{-1}mol^{-1}\times 300k}{2.57\times 10^{-3}bar\times 0.200L}$
$\Rightarrow 61039gmol^{-1}$
Hence (A) is the correct answer.
answered Jun 10, 2014 by sreemathi.v

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