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The solubility of oxygen in water is $2.43\times 10^{-3}$ mole/litre at $25^{\large\circ}$C and 1.0atm pressure.Calculate the concentration of oxygen at $25^{\large\circ}$C and 0.2atm pressure.

$\begin{array}{1 1}4.86\times 10^{-4}mole\\3.86\times 10^{-4}mole\\2.86\times 10^{-4}mole\\1.86\times 10^{-4}mole\end{array} $

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According to Henry's law
$C=K_H P$
$K_H$=Hentry's law constant
Hence $K_H=\large\frac{C}{P}$
$\Rightarrow \large\frac{2.43\times 10^{-3}}{1.0}$
$\Rightarrow 2.43\times 10^{-3}mol lit ^{-1} atm^{-1}$
Further $C=K_H P$
Or C=$2.43\times 10^{-3}\times 0.2$
Concentration =$4.86\times 10^{-4}$ mole
answered Aug 6, 2014 by sreemathi.v

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