if N2 gas is bubbled through water at 293 K how much percentage of N2 will be present in the solution if and to exert partial pressure of 0.9 bar given ke of N2 at 293 K is 76 kilo bar.
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According to Henry's law,
pN2=KH.XN2
⇒xN2=768000.987=1.29×10−5
Moles of water in 1L=181000=55.56
Let us take the moles of nitrogen as n
Total moles =n+55.56
∴ xN2=n+55.56n=1.29×10−5
⇒n(1−1.29×10−5)−55.56×1.29×10−5=0
⇒n=0.7×10−3=0.7 milimoles