The Henry's law constant fo 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 298K and 5 atm pressue is

Options

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

Answer

A

Explanation

`P_(N_(2))=K_(H)xx X_(N_(2))`
`X_(N_(2))=(1)/(10^(5))xx0.8xx5=4xx10^(-5)` per mole
In 10 mole, solubility is `4xx10^(-4)`.