The Henry's law constant for oxygen dissolved in water is 4.34 xx10^(4) atm at 25^(@)C. If the the partial pressure of oxygen in air is 0.2 atm under atmospheric conditions, calculate the concentration (in moles per litre) of dissolved oxygen in water in equilibrium with air at 25^(@)C.

The Henry's law constant for oxygen dissolved in water is 4.34 xx10^(4

1.	The Henry's law constant for oxygen dissolved in water is 4.34 xx10^(4) atm at 25^(@)C. If the the partial pressure of oxygen in air is 0.2 atm under atmospheric conditions, calculate the concentration (in moles per litre) of dissolved oxygen in water in equilibrium with air at 25^(@)C.
Answer» Solution :`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)`