Dry air contains 79 % ,N_2 and 21 % ,O_2. At | vedclass

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(C) Total pressure of air over water $= 1 \,atm$. Partial pressure of $N_2$ and $O_2$ are: $P_{N_2} = \frac{1 	imes 79}{100} = 0.79 \,atm$ $P_{O_2} =

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$2: 1$

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(C) Total pressure of air over water $= 1 \,atm$. Partial pressure of $N_2$ and $O_2$ are: $P_{N_2} = \frac{1 	imes 79}{100} = 0.79 \,atm$ $P_{O_2} = \frac{1 	imes 21}{100} = 0.21 \,atm$ Applying Henry's law $(P = K_H 	imes X)$: $X_{N_2} = \frac{P_{N_2}}{K_{H, N_2}} = \frac{0.79}{8.57 	imes 10^4} \approx 9.22 	imes 10^{-6}$ $X_{O_2} = \frac{P_{O_2}}{K_{H, O_2}} = \frac{0.21}{4.56 	imes 10^4} \approx 4.60 	imes 10^{-6}$ Ratio of mole fractions of $N_2$ and $O_2$: $\frac{X_{N_2}}{X_{O_2}} = \frac{9.22 	imes 10^{-6}}{4.60 	imes 10^{-6}} \approx 2$ Thus, the ratio is $2: 1$.

Gas	$K_{H} / \text{kbar}$
$Ar$	$40.3$
$CO_{2}$	$1.67$
$HCHO$	$1.83 \times 10^{-5}$
$CH_{4}$	$0.413$

Gas	$K_H$ (kbar at $298 \ K$)
$CO_2$	$1.67$
$Ar$	$40.3$
$HCHO$	$1.83 \times 10^{-5}$
$CH_4$	$0.413$

Dry air contains $79 \% \,N_2$ and $21 \% \,O_2$. At $T \,K$, if Henry's law constants for $N_2$ and $O_2$ in water are $8.57 \times 10^4 \,atm$ and $4.56 \times 10^4 \,atm$ respectively, the ratio of mole fractions of $N_2$ and $O_2$ dissolved in water at $1 \,atm$ is

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