The Henry's Law constant for oxygen dissolved in water is ` 4.34 xx 10 ^(4)atm ` ( mol /L) at ` 25^(@) C ` . If the partial pressure of oxygen in air is 0.2 atm under ordinary atmospheric condition calculate the concentration (in mol/litre ) of dissolved oxygen in water equilibrium with air at ` 25^(@) C `

Explanation

` K_H " " =4.34 xx 10_ 4 atm//(mol//L) `
`P+_(O_2) " " =0.2atm `
` T" " =25+ 273 = 293K.`
`C prop P `
C=KP
P= ` K_H xx x`
` x = (P)/( K_H) =( 0.2 atm )/( 4.34xx 10 ^(4) atm (mol xxL)) `
`= 0.04 xx 10 ^(-4) mol //L `