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The Henry's Law constant for oxygen diss...

The Henry's Law constant for oxygen dissolved in water is `4.34xx10^(4) atm at 25^(@)C`. If partial pressure of exygen in are is 0.2 atm. Under ordinary atmospheric conditions, calculate the concentration (in moles/litre) of dissolved oxygen in water in equilibrium with air at `25^(@)C`.

Text Solution

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`"Mole fraction of "O_(2)(x_(o)_(2))=("Partial pressure of" O_(2))/(K_(H)"for" O_(2))=((0.2 atm))/((4.34xx10^(4)atm))=4.6xx10^(-6)`
`x_(o_(2))=n_(o_(2))/(n_(o_(2))+n_(H_(2))o)=n_(o_(2))/(n_(H_(2))o)`
`(n_(o_(2))="has been neglected in this expression in this expression since the gas is very little soluble in water")`
`x_(o_(2))=4.6xx10^(-6),((1000 g))/((18 g mol^(-1)))=55.55 mol.`
`n_(o_(2))= (x_(o_(2)))xxn_(H_(2)o)=(4.6xx10^(-6))xx(55.55mol)=(2.55 mol)=2.55xx10^(-4)mol`
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