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The Henry's law constant for the solubli...

The Henry's law constant for the solublity of `N_(2)` gas in water at `298K` is `1.0xx10^(5)` atm. The mole fraction of `N_(2)` air is `0.8`. The number of moles of `N_(2)` from air dissolved in `10` moles of water of `298K` and `5` atm pressure is:

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The correct Answer is:
4
4
`X_(N_(2))=(P_(N_(2)))/(K_(H))=(5xx0.8)/(1xx10^(5)) (P_(N_(2))=P_("air")xx "mole fraction of" N_(2))`
`:. (n_(N_(2)))/(n_(N_(2))+n_(H_(2)O))=4xx10^(-5)=(n_(N_(2)))/(n_(N_(2))+10) (n_(N_(2))lt lt lt10)`
`:.n=4xx10^(-4)`
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