The henry's law constant for 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 298 K and 5 atm pressure is

Options

A. `4.0xx10^(-4)`
B. `4.0xx10^(-5)`
C. `5.0xx10^(-4)`
D. `4.0xx10^(-6)`

Explanation

Henry's law, `(2.0)=(2.42)=4.82-P_("partial")=P_(H)=(X_N_(2)))_("in solution")` pressure of `N_(2)`
`underset(darr)(5)xxunderset(darr)(0.8)=10^(5)xxx_(N_(2))` in solution
`P_("Tota,l")(X_(N_(2)))` in air
`impliesX_(N_(2))=4xx10^(-5)=(n_(N_(2)))/(n_(N_(2))+n_(0_(2)))=(n_(N_(2)))/(n_(0_(2)))iimpliesn_(N_(2))=4xx10^(-4)`