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77.7 k bar Solution :

Mass of one litre = 1000 g <br> No. of moles of water `(m_(1))` in one litre `=(1000)/(18)=55.5` <br> Mass of `N_(2)=2xx10^(-5) kg =2xx10^(-2)g` <br> Gram mol. mass of `N_(2)=2xx14=28"g mol"^(-1)` <br> No. of moles of nitrogen `(n_(2))=(2xx10^(-2))/(28)=7.14 xx 10^(-4)` <br> Mole fraction of `N_(2)` <br> `=(n_(2))/(n_(1)+n_(2))=(7.14 xx 10^(-4))/(55.5+7.14xx10^(-4))=1.286xx10^(-5)` <br> According to Henry.s law, `p_(N_(2))=` Henry.s law constant `(K_(H)) xx` Mole fraction of `N_(2)` <br> Substituting the values, 1 bar `=K_(H) xx 1.286 xx 10^(-5)` <br> or `K_(H)=(1"bar")/(1.286xx10^(-5))=77760` bar <br> `=77.7` K bar