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At 300K,36g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar.If osmotic pressure of solution is 1.52 bar at the same temperature,what would be its concentration?

$\begin{array}{1 1}3.28L\\3.18L\\2.28L\\2.18L\end{array} $

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1 Answer

$\pi=\large\frac{n}{V}$$RT$
$\pi V=nRT$
If n,R and T are same in both case.
$\pi_2V_2=\pi_1V_1$
$V_2=\large\frac{\pi_1V_1}{\pi_2}$
$\;\;\;\;=\large\frac{4.98\times 1}{1.52}$
$\;\;\;\;=3.28L$
answered Aug 8, 2014 by sreemathi.v
 
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