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

According to Henry's law
`x_(N_2)xxK_H =p_(N_2)` ( `p_(N_2)` = Partial pressure of `N_2`)
Given, total pressure = 5 atm
mole fraction of `N_2`=0.8
`therefore` Partial pressure of `N_2`=0.8 x 5 =4
`rArr x_(N_2)xx1 xx10^5 = 4 rArr x_(N_2)=4 xx10^(-5)`
no. of moles of `H_2O, n_(H_2O)=10`
no. of moles of `N_2, n_(N_2)`=?
`n_(N_2)/(n_(N_2)+n_(H_2O))=x_(N_2) = 4 xx10^(-5) rArr n_(N_2)/(10+n_(N_2))=4xx10^(-5)`
`rArr n_(N_2)=4xx10^(-4) ` [`because n_(N_2)` lt lt lt 10 ]