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, from air dissolved in 10 moles of water at 298 K and 5 atm pressure is

Options

A. `4.0 xx 10^(-4)`
B. `4.0 xx 10^(-5)`
C. `5.0 xx 10^(-4)`
D. `4.0 xx 10^(-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`
`:. " Partial pressure of " N_2 = 0.8 xx 5 = 4`
`implies x_(N_2) xx 1 xx 10^5 = 4 implies x_(N_2) implies x_(N_2) =4 xx 10^(-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) = 4xx10^(-5) implies (n_(N_2))/(10+n_(N_2))= 4xx10^(-5)`
`implies n_(N_2)=4xx10^(-4) " " [ :. n_(N_2) lt lt lt 10]`