`N_(2)` gas is bubbled through water at `293 K` and the partial pressure of `N_(2)` is `0.987` bar .If the henry's law constant for `N_(2)` at `293 K` is `76.84` kbar, the number of millimoles of `N_(2)` gas that will dissolve in `1 L` of water at `293 K` is

Options

A. `1.29`
B. `0.716`
C. `2.29`
D. `7.16`

Answer

b

Explanation

`P_(N_2) = K_(H) + X_(O_2)`
`X_(O_3) =- (P_(O_2))/(K_(H)) = (0.987)/(76.48 + 10^(3)) = 1.29 + 10^(-5)`
`n_(N_2) = 129 + 10^(-3) xx 55.5 = 0.716 mmol`