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. `4.0 xx 10^(-4) atm`
B. `4.0 xx 10^(-5) atm`
C. `5.0 xx 10^(-4) atm`
D. `4.0 xx 10^(-6) atm`

Answer

A

Explanation

`P_(N2)=5 xx 0.8 =4`
Applying Henry's law, we get
`P_(N2)=K_(H) xx chi_(N_(2))` or `4=10^(5) xx chi_(N_(2))`
chi_(N_(2))=4 xx 10^(-5)`
`n_("nitrogen")/(n_("nitrogen")+n_(H_(2)O))=4 xx 10^(-5)`
or `n_("nitrogen")/(n_("nitrogen")+10)=4 xx 10^(-5)`
or `n_("nitrogen")=4 xx 10^(-4)`.