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Determine the osmotic pressure of a solution prepared by dissolving 25 mg of $K_2SO_4$ in 2 litre of water at 25$^{\large\circ}$ C, assuming that it is completely dissociated.

$\begin{array}{1 1}5.27\times 10^{-3}atm\\6.27\times 10^{-3}atm\\4.27\times 10^{-3}atm\\3.27\times 10^{-3}atm\end{array} $

1 Answer

$K_2SO_4$ dissolved =25mg =0.025g
Volume of solution =2 litre
Molar mass of $K_2SO_4=2\times 39+32+64$
$\Rightarrow 174gmol^{-1}$
As $K_2SO_4$ dissociates completely as
$K_2SO_4\rightarrow 2K^++SO_4^{2-}$
(ie) ions produced (i)=3
$\pi =i\times \large\frac{w}{M}\times \frac{1}{V}$$RT$
$\Rightarrow 3\times \large\frac{0.025g}{174gmol^{-1}}\times \frac{1}{2L}$$\times 0.821 L atm K^{-1}mol^{-1}\times 298 K$
$\Rightarrow 3\times \large\frac{0.025g}{174gmol^{-1}}\times \frac{1}{2L}$$\times 0.821Latm K^{-1}mol^{-1} \times 298 K$
$\Rightarrow 5.27\times 10^{-3}$ atm
answered Aug 6, 2014 by sreemathi.v
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