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Q.

The 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

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a

$4.0 \times 10^{- 4}$

b

$4.0 \times 10^{- 5}$

c

$5.0 \times 10^{- 4}$

d

$4.0 \times 10^{- 6}$

answer is A.

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Detailed Solution

$\begin{array}{l} P = K_{H} χ_{N_{2}} \\ 0.8 \times 5 = 1 \times 10^{5} \times χ_{N_{2}} \\ χ_{N_{2}} = 4 \times 10^{- 5} (in 10 moles of water) \\ \Rightarrow 4 \times 10^{- 5} = \frac{n_{N_{2}}}{n_{N_{2}} + 10} \\ n_{N_{2}} \times 5 \times 10^{- 5} + 4 \times 10^{- 4} = n_{N_{2}} \\ \Rightarrow n_{N_{2}} = 4 \times 10^{- 4} \end{array}$

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The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0×105atm . 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