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The osmotic pressure of a solution $(density =1.02\;g \;cm^{-3})$ Containing 50 g glucose $(C_6H_{12}O_6)$ in 1 kg of water at 300 K is

$\begin{array}{1 1} 67.39\;kPa \\673.85\;kPa \\ 6.74\;kPa \\ 673.85\;Pa \end{array} $

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Answer :$673.85\;kPa$
Concentration of glucose $ = \large\frac{(50/180)\;mol}{(1.05\;kg /1.02 kg L^{-1})}$$=0.27 \;mol \;L^{-1}$
Osmotic pressure $\pi =cRT=(0.27 \;mol\;L^{-1})(8.314 kPa\;L\;K^{-1}\;mol^{-1})(300 K) =673.85 \;kPa$
answered Aug 7, 2014 by meena.p

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