SolutionHard
Question
The Henry’s law constant for the solubility of N2 gas in water at 298 K 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 298 K and 5 atm pressure is
Options
A.4.0 × 10−4
B.4.0 × 10−5
C.5.0 × 10−4
D.5.0 × 10−5
Solution
$P_{N_{2}} = K_{H}.X_{N_{2\left( \text{solution} \right)}}$
Or, $5 \times 0.8 = \left( 1.0 \times 10^{5} \right) \times \frac{n_{N_{2}}}{n_{N_{2}} + 10} \simeq 10^{5} \times \frac{n_{N_{2}}}{10}$
$\therefore n_{N_{2}} = 4 \times 10^{- 4}$
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