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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 `298K` is `1.0 xx 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 298K and 5 atm pressure is

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Accoding to Henry's Law , `P_(N_(2))=K_(H)x_(N_(2))`
`x_(N_(2))=P_(N_(2))/K_(H)=((4.0 atm))/((1.0xx10^(5)atm))=4xx10^(-5)`
`n_(N_2)/(n_(N_2)+n_(H_2)=x_(N_(2))=4xx10^(-5)`
`n_(N_2)/n_(H_2)=x_(N_(2))=4xx10^(-5)`
`n_(N_2)=4xx10^(-5)xxn_(H_2)o=4xx10^(-5)xx(10 mol)`
`4xx10^(-4)mol.`
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