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. `4.0xx10^(-4)`
B. `4.0xx10^(-5)`
C. `5.0xx10^(-4)`
D. `4.0xx10^(-6)`

Answer

(A)

Explanation

`P=K_(H)chi_(N_(2))`
`0.8xx5=1xx10^(5)xx chi_(N_(1))`
`chi_(N_(2))=4xx10^(-5)` (in 10 moles of water)
`rArr 4xx10^(-5)=(n_(N_(1)))/(n_(N_(1))+10)`
`n_(N_(1))xx5xx4xx10^(-4)=n_(N_(2))`
`rArrn_(N_(1))=4xx10^(-4)`