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` moles of water at `298 K` and `5 atm`. Pressure is:

Options

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

Answer

A

Explanation

`p_(N_(2))=x_(N_(2))xxp_("total")`
`=0.8xx5=4` atm
According to Henry's law :
`p_(N_(2))=K_(H)xxx_(N_(2))`
`4=1xx10^(5)xx(n_(N_(2)))/(n_(N_(2))+n_(H_(2)O))`
`(n_( N_(2))+n_(H_(2)O))/(n_(N_(2)))=(1xx10^(5))/(4)`
`n_(N_(2))=4xx10^(-4)`]