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Questions  >>  CBSE XII  >>  Chemistry  >>  Solutions
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The Henry's law constant for oxygen dissolved in water is $4.34\times 10^4$ atm at $25^{\large\circ}$C.If the partial pressure of oxygen in air is 0.2atm,under atmospheric conditions,calculate the concentration (in moles per litre) of dissolve oxygen in water in equilibrium with air at $25^{\large\circ}$C.

$\begin{array}{1 1}2.55\times 10^{-4}M\\3.55\times 10^{-4}M\\4.55\times 10^{-4}M\\1.55\times 10^{-4}M\end{array} $

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