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The henry's law constant for the solubil...

The henry's law constant for the solubility of `N_2` gas in water at 298 K is `1.0xx10^5` 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. Partial pressure of `N^(2)` in air `(p_(N_2)) = P_("total") xx (chi_(N_2))("in air")`
`=5 xx 0.6`
`P_(N_2)`("in air" ) `= K_H xx chi_(N_2)("in" (H_(2)O))`
`5 xx 0.6 = 1.0 xx 10^(5) xx chi^(N_(2))("in" H_(2)O)`
`chi_(N_(2))` in 10 moles of water `=(5 xx 0.6)/(1.0 xx 10^(5)) = 3.0 xx 10^(-5)`
`chi_(N_(2))= n_(N_(2))/(n_(N_(2)) + n_(H2O))`
`3.0 xx 10^(-5) = n_(N_(2))/(n_(N_(2)) + 10)`
`n_(N_(2)) xx 3 xx10^(-5) xx 10 = n_(N_(2))`
`3 xx 10^(-4) = n_(N_(2)) (1-3 xx 10^(-5)) [1-3 xx 10^(-5) = 1]`
`:. n^(N_(2)) = 3 xx 10^(-4)`
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