$\begin{array}{1 1}30kgmol^{-1}\\40kgmol^{-1}\\60kgmol^{-1}\\80kgmol^{-1}\end{array} $

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Answer : $60kgmol^{-1}$

The osmotic pressure of solution becomes

$\Pi=h\rho g=(83.1\times 10^{-3}m)(10^3kg m^{-3})(10m s^{-2})=831kg m^{-1}s^{-2}=831Pa$

Using the expression $\Pi=cRT=\large\frac{m/M}{V}$$RT$,we get

$831 Pa=\big(\large\frac{20g}{M}\frac{1}{dm^3}\big)$$(8.314\times 10^3Pa \;dm^{3}K^{-1}mol^{-1})(300K)$

$M=\big(\large\frac{20g\times 8.314\times 10^3\times 300}{831}\big)$$gmol^{-1}=60,000gmol^{-1}=60kgmol^{-1}$

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