The Henry's law constant for the solubility of Nitrogen gas iin water at 350 K is `8xx10^4` atm. The mole fraction of nitrogen in air is 0.5 . The number of moles of Nitrogen from air dissolved in 10 moles of water at 350 K and 4 atm pressure is …………….... .

Options

A. `4xx10^(-4)`
B. `4xx10^(4)`
C. `2xx10^(-2)`
D. `2.5 xx 10^(-4)`

Answer

D

Explanation

`K_(H) = 8xx10^(4)`
`(x_(N_(2)))_("in air")= 0.5`
Total pressure = 4 atm
Partial pressure of nitrogen = Mole fraction`xx` Total pressure
=`0.5xx4 = 2`
`(P_(N_(2))) = K_(H)xx "Mole fraction of" N_(2) "in solution"`
`2=8xx10^(4)xx("Number of moles of nitrogen")/("Total number of moles")`
`(10+"No. of moles of" N_(2))/("No. of moles of" N_(2)) = (8xx10^(4))/(2)`
`(10)/("No. of moles of" N_(2))+1 = 4xx10^(4)`
`(10)/("No. of moles of" N_(2)) = 40000-1`
`therefore` No. of moles of `N_(2) = (10)/(39999) = 2.5xx10^(-4)`