Mole fraction of N2 gas in air is 0.8. Henry law constant for N2 in water is 2.22×104 atm. The moles of N2 gas present in 1 L of solution which is in equilibrium with air at 10 atm is a×10−2 . Find a?

Pgas=KH.Xgas Pgas=10×0.8 =8atm8 =2.22×104 Xgas Xgas=82.22×104=3.6×10−4ngasnwater=3.6×10−4ngas=3.6×10−4×55.56 =200×10−4 =2×10−2

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Mole fraction of $N_{2}$ gas in air is 0.8. Henry law constant for $N_{2}$ in water is $2.22 \times 10^{4}$ atm. The moles of $N_{2}$ gas present in 1 L of solution which is in equilibrium with air at 10 atm is $a \times 10^{- 2}$ . Find a?

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Detailed Solution

$P_{g a s} = K_{H} . X_{g a s} \begin{array}{l} P_{g a s} = 10 \times 0.8 \\ = 8 a t m \\ 8 = 2.22 \times 10^{4} X_{g a s} \end{array} \begin{array}{l} X_{g a s} = \frac{8}{2.22 \times 10^{4}} = 3.6 \times 10^{- 4} \\ \frac{n_{g a s}}{n_{w a t e r}} = 3.6 \times 10^{- 4} \\ n_{g a s} = 3.6 \times 10^{- 4} \times 55.56 \\ = 200 \times 10^{- 4} \\ = 2 \times 10^{- 2} \end{array}$

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