The Henry’s law constant for $O_{2}$ dissolved in water is $4.34 \times 10^{4}$ atm at certain temperature. If the partial pressure of $O_{2}$ in a gas mixture that is in equilibrium with water is 0.434 atm, what is the mole fraction of $O_{2}$ in solution?

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$1 \times 10^{- 4}$

$2 \times 10^{- 6}$

$1 \times 10^{- 5}$

$2 \times 10^{- 5}$

answer is A.

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Detailed Solution

According to Henry’s law, $P = K_{H} \times x$

$X_{O_{2}} = \frac{P_{O_{2}}}{K_{H}} = \frac{0.434}{4.34 \times 10^{4}} = 1 \times 10^{- 5}$

Mole fraction of $O_{2}$ in water $= 1 \times 10^{- 5} .$

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