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The Henry's law constant for the solubility of \(\mathrm{N}_2\) gas in water at \(298 \mathrm{~K}\) is \(1 \times 10^{+5} \mathrm{~atm}\). The mole fraction of air is 0.8 . The number of moles of \(\mathrm{N}_2\) from air dissolved in 10 moles of water at \(298 \mathrm{~K}\) and 5 atm pressure is
- A \(4 \times 10^{-5}\)
- B \(4 \times 10^{-4}\)
- C \(5 \times 10^{-4}\)
- D \(4 \times 10^{-6}\)
Answer & Solution
Correct Answer
(B) \(4 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
At total pressure \(5 \mathrm{~atm}\), the partial pressure of \(\mathrm{N}_2=5 \times 0.8=4 \mathrm{~atm}\) According to Henry's Law, \[ \mathrm{P}_{\mathrm{N}_2}=\mathrm{K}_{\mathrm{H}} \mathrm{x}_{\mathrm{N}_2} \] \(\left(\mathrm{x}_{\mathrm{N}_2}\right)\) is the mole…
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