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Question

Henry's law constant for the solubility of $N_2$ gas in water at 298 K is $1.0 \times 10^5$ atm. The mole fraction of $N_2$ in air is 0.8. The number of moles of $N_2$ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is

$4.0 \times 10^{-4}$

$4.0 \times 10^{-5}$

$5.0 \times 10^{-5}$

$4.0 \times 10^{-6}$

Solution

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