The Henry's law constant fo 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 298K and 5 atm pressue is

Explanation

`P_(N_(2))=K_(H)x_(N_(2))`
`P_(N_(2))` = mole fraction of `N_(2)` in air `xx` Total pressure of air `= 0.8xx5` atm = 4 atm
`therefore 4 atm = 10^(5)atm xx x_(N_(2))` or `x_(N_(2))=4xx10^(-5)`
This means that in one mole of water, no. of moles of `N_(2)` dissolved `= 4xx10^(-5)`
`therefore` In 10 moles of water, no. of moles of `N_(2)` is dissolved `= 4xx10^(-4)`