1 kg of water under a nitrogen pressure of 1 atmosphere dissolves 0.02 gm of nitrogenat 293 k. Calculate Henry' s law constant :

Answer

`7.75 xx 10^(4)` atm

Explanation

`p= K_(H).x`
`x = ("Moles of " N_(2))/("Moles of" N_(2) + "Moles of " H_(2)O)`
`= ((0.02)/(28))/((0.02)/(28)+(1000)/(18)) = 7.75 xx 10^(4) atm`
` p=1 atm`
`K_(H) = (p)/(x)`
` = (1)/(1.129 xx 10^(-5))=7.75 xx 10^(4) atm`