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Henry 's law constant for the solubility...

Henry 's law constant for the solubility of `N_(2)` gas in water at `298 K` is `1.0 xx 10^(5) atm`. The mole fraction of `N_(2)` in air is `0.6`. The number of moles of `N_(2)` from air dissolved in 10 moles of water at `298 K`and `5 atm` pressure is
a. `3.0 xx 10^(-4)` b. `4.0 xx 10^(-5)` c. `5.0 xx 10^(-4)` d.`6.0 xx 10^(-6)`

Text Solution

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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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