The Henry's law constant for the solubility of `N_2` gas in water at 298 K 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 mole of water at 298 K and 5 atm pressure is:

Options

A. `4.0xx10^(-4)`
B. `4.0 xx 10^(-8)`
C. `5.0xx 10^(-4)`
D. `4.0 xx 10^(-6)`

Answer

A

Explanation

`P.= P_(T) X_(N_(2))`
`therefore P_(N_(2)) =5 xx 0.8= 4.0` atm
From Henry.s law `P_(N_(1))= K_(H).X_(N_(2))` dissolved
`therefore X_(N_(2))= (4)/(1 xx 10^(5)) =4 xx 10^(-5) or (n_(N_(2)))/(n_(N_(2)) + N_(H_(2)O))= (n_(N_(2)))/(n_(N_(2)) + 10)= 4 xx 10^(-5)`
`therefore n_(N_(2))= 4 xx 10^(-5) xx 10= 4 xx 10^(-4)`