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

Options

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

Explanation

`P_(N_(2)) = K_(H) xx x_(N_(2))`
`x_(N_(2)) = (1)/(10^(5)) xx 0.8 xx 5 = 4 xx 10^(-5)` per mole
In 10 mole solubility is `4xx 10^(-4)`.