The Henry’s law constant for solubility of N2 gas in water at 298k is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298k and 5 atm pressure is:

5×0.81×105=nN210;nN2=4×10×10−5=4×10−4

The Henry’s law constant for solubility of N2 gas in water at 298k is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298k and 5 atm pressure is:

5×0.81×105=nN210;nN2=4×10×10−5=4×10−4

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The Henry’s law constant for solubility of N₂ gas in water at 298k is 1.0 × 10⁵ atm. The mole fraction of N₂ in air is 0.8. The number of moles of N₂from air dissolved in 10 moles of water at 298k and 5 atm pressure is:

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4 × 10^–4

4.0 × 10^–6

4.0 × 10^–5

5.0 × 10^–4

answer is A.

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$\frac{5 \times 0 .8}{1 \times 10^{5}} = \frac{n_{N_{2}}}{10}; n_{N_{2}} = 4 \times 10 \times 10^{- 5} = 4 \times 10^{- 4}$

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The Henry’s law constant for solubility of N2 gas in water at 298k is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298k and 5 atm pressure is:

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The Henry’s law constant for solubility of N₂ gas in water at 298k is 1.0 × 10⁵ atm. The mole fraction of N₂ in air is 0.8. The number of moles of N₂from air dissolved in 10 moles of water at 298k and 5 atm pressure is: