The Henry's law constant for the solubility of Nitrogen gas in water at ` 350 K` is ` 8xx 10^(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. `4 xx 10 ^(-4)`
B. `4 xx 10 ^(x)`
C. `2xx 10 ^(-2)`
D. `2.5 xx 10 ^(-4)`

Answer

D

Explanation

`K _(H) = 8 xx 10 ^(4)`
`(X_(N_(2))) _("in air") = 0.5`
Total pressure = 4 atm
Partial pressure of nitrogen = Mole fraction `xx` Total pressure
`0.5 xx 4 =2`
`(P _(N _(2))) = K _(H) xx` Mole fraction of `N_(2)` in solution
`2 = 8 xx 10 ^(4) xx ("Number of moles of nitrogen")/("Total number of moles")`
`(10 + "No. of moles of" N _(2))/( "No. of moles of" N _(2)) = (8xx 10 ^(4))/(2)`
`(10)/("No. of moles of" N_(2)) +1 = 4 xx 10 ^(4)`
`(10)/("No. of moles of" N_(2)) = 4000 -1`
`therefore ` No. of moles of `N _(2) = (10)/(39999) =2.5 xx 10 ^(-4)`