Applying Henry's law: The solubility of pure `N_(2)(g)` at `25^(@)C` and 1 atm is `6.8xx10^(-4) mol L^(-1)`. If the partial pressure of `N_(2)(g)` in the atmosphere is `0.78 atm`, calculate the concentration of `N_(2)(g)` dissolved in water under atmospheric conditions.
Strategy: According to Henry's law, the solubility of a gas in water is `KP`. Thus, the first step is to calculate the quantity `K`.

Explanation

According to Equation `(2.41)`
Solubility `(S)= KP`
`6.8xx10^(-4) mol L^(-1)= K(1 atm)`
`K= 6.8xx10^(-4) mol L^(-1) atm^(-1)`
Therefore, the solubility of `N_(2)(g)` in water is
`S=(6.8xx10^(-4) mol L^(-1) atm^(-1))(0.78 atm)`
`=5.3xx10^(-4) mol L^(-1)`
`=5.3xx10^(-4) M`