The Henry's Law constant for oxygen dissolved in water is `4.34xx10^(-4) Catm^(-1)` 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`.

Answer

`2.55xx10^(-4)"mol L"^(-1)`

Explanation

`P_(O_(2))=K_(H)xxx_(O_(2))" or "x_(O_(2))=(P_(O_(2)))/(K_(H))=("0.2 atm")/(4.34xx10^(4)" atm")=4.6xx10^(-6)`
`"To convert it into molarity. "x_(O_(2))=(n_(O_(2)))/(n_(O_(2))+n_(H_(2)O))`
`"For 1 litre of water, "m_(H_(2)O)=1000//18="55.5 moles "therefore (m_(O_(2)))/(55.5)=4.6xx10^(-6)" or "n_(O_(2))=2.55xx10^(-4)" mole"`
`"Hence, molarity "=2.55xx10^(-4)"mol L"^(-1)`