Henry law constant for oxygen dissolved in water is` 4.34 xx 10^(4) atm` at `25^(@) C`. If the partial pressure of oxygen in air is 0.4 atm.Calculate the concentration ( in moles per litre) of the dissolved oxygen in equilbrium with air at `25^@C`.

Explanation

Given `K_(11)=4.34xx10^4 atm`
`Po_2=0.4 atm`
acc. to Henry's Law
`p=K_(H)X`
`X_(O_2)=0.4/(4.34xx10^4)=9.2xx10^(-6)`
Moles of water `(n_(H_2O))=1000/18=55.8 ` mol
`X_(O_2)=n_(O_2)/(n_(O_2)+0)`
or `X_(O_2)xxn_(H_2O)=n_(O_2)`
So `n_(O_2)=9.2xx10^(-6)xx55.55=5.11xx10^(-4)mol`
So `M=5.11xx10^(-4)`