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(D) According to Henry's Law,$P = K_H \cdot x$,where $x$ is the mole fraction of $O_2$ in water.$20 \ kPa = (8.0 	imes 10^{4} \ kPa) \cdot x$$x = \fr

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$25$

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(D) According to Henry's Law,$P = K_H \cdot x$,where $x$ is the mole fraction of $O_2$ in water.$20 \ kPa = (8.0 	imes 10^{4} \ kPa) \cdot x$$x = \frac{20}{8.0 	imes 10^{4}} = 2.5 	imes 10^{-4}$.Since $x = \frac{n_{O_2}}{n_{O_2} + n_{H_2O}} \approx \frac{n_{O_2}}{n_{H_2O}}$ (as $n_{O_2} \ll n_{H_2O}$).For $1 \ dm^{3}$ of water,mass $= 1000 \ g$,so $n_{H_2O} = \frac{1000}{18} = 55.55 \ mol$.$n_{O_2} = x \cdot n_{H_2O} = (2.5 	imes 10^{-4}) 	imes 55.55 = 1.38875 	imes 10^{-2} \ mol$.This is the molar solubility in $mol \ dm^{-3}$,which is $1388.75 	imes 10^{-5} \ mol \ dm^{-3}$.Rounding to the nearest integer,we get $1389$.

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