Question

let the number of mole of ( mathrm{N} 2 ) dissolved in water be ( mathrm{n} )
Henry's law constant for N 2 at ( 293 mathrm{K} ) is ( 76.48 mathrm{kbar} ).
according ti henrys law
partial pressure exerted by ( mathrm{N} 2=mathrm{Kh}^{*} mathrm{x} ) ( x ) is the mole fraction
mole fraction of ( mathrm{N} 2=x /(mathrm{x}+55.5) ) ( .987=76.48^{*} 10^{wedge} 3^{*} x /(x+55.5) )
( mathrm{x}=.7 mathrm{milimole} )

# nitrogen gas N, is passed in water at 293 K, then how many milli mole N2 gas will be soluble in one litre water? The partial pressure of N2 is 0.987 bar and the value of Henry's law constant Ku at 293K is 76.48 bar.

Solution