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. `4 xx 10^(-4)`
B. `4.0 xx10^(-5)`
C. `5 xx 10^(-4)`
D. `4.0 xx 10^(-6)`

Answer

A

Explanation

`p_(N_(2))=K_(H)xx x_(N_(2))`
`y(N_(2))=(p_(N_(2)))/(P_("Total"))`
`:. p_(N_(2))=P_("Total")xxy(N_(2))`
`= 5 atm xx 0.8 =4` atm
`:. p_(N_(2))=K_(H)xxxN_(2)`
`4=1.0xx10^(5)xx x_(N_(2))`
`x_(N_(2))=4xx10^(-5)`
`(x_(N_(2)))/(n_(N_(2))+n_(H_(2)O))=4xx10^(-5)`
or`n_(N_(2))=4xx10^(-4)" "(becausen_(H_(2)O)=10)`.