The Henry's law constant for the solubility of `N_(2)` gas i water at `298K` is `1xx10^(5)` atm. The mole fraction of `N_(2)` air is `0.8`. The number of mole of `N_(2)` of dissolved in 10 mole of water at `298K` and `5` atm. are `x xx10^(4)`. Find the value of `x`.

Answer

4

Explanation

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)`