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A solution containing $8.6gL^{-1}$ of urea (molar mass $60gmol^{-1}$) is isotonic with a 5% solution of unknown solute.The molar mass of the solute will be

$\begin{array}{1 1}348.9gmol^{-1}\\174.5gmol^{-1}\\87.3gmol^{-1}\\34.89gmol^{-1}\end{array} $

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Answer : $348.9gmol^{-1}$
Isotonic solution has the same osmotic pressure and hence the same concentration in the solution.
$\large\frac{(8.6g/60gmol^{-1})}{1L}=\frac{(5g/M)}{0.1L}$
Hence $M=\large\frac{5\times 60}{8.6\times 0.1}$$gmol^{-1}=348.8gmol^{-1}$
answered Jul 23, 2014 by sreemathi.v
 

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