If O_2 gas is bubbled through water at 303 ; | vedclass

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(B) According to Henry's Law,$p = K_H 	imes x$,where $x$ is the mole fraction of the gas in the solution.Given $p = 0.920 \; bar$ and $K_H = 46.82 \;

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$1$

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(B) According to Henry's Law,$p = K_H 	imes x$,where $x$ is the mole fraction of the gas in the solution.Given $p = 0.920 \; bar$ and $K_H = 46.82 \; kbar = 46820 \; bar$.$x = \frac{p}{K_H} = \frac{0.920}{46820} \approx 1.965 	imes 10^{-5}$.Since the solubility is very small,the number of moles of $H_2O$ in $1 \; L$ $(1000 \; g)$ is $n_{H_2O} = \frac{1000}{18} \approx 55.55 \; mol$.$x = \frac{n_{O_2}}{n_{O_2} + n_{H_2O}} \approx \frac{n_{O_2}}{n_{H_2O}}$.$n_{O_2} = x 	imes n_{H_2O} = (1.965 	imes 10^{-5}) 	imes 55.55 \approx 1.09 	imes 10^{-3} \; mol$.$n_{O_2} \approx 1.09 \; millimoles$.The nearest integer is $1$.

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