At what partial pressure , oxygen will have a solubility of `0.06 gL^(-1)` in water at 298 K ? Henry's law constant `(K_(H))` of `O_(2)` in water at 303 K is 46.82 k bar .(Assume the density of the solution to be the same as that of water).

Answer

1.58 bar

Explanation

Mass of 1L solution = 1000g
Mass of water ` = 1000-0.06 = 999.94g`
Moles of water ` = (999.94)/(18) = 55.55`
Moles of `O_(2) = (0.06)/(32) = 1.875 xx 10^(-23)`
`therefore ` Mole fraction of `O_(2) ,x_(O_(2)) = (1.875 xx 10^(-3))/(55.55 + 1.875 xx 10^(3))`
`= (1.875 xx 10^(-3))/(55.55) = 3.37 xx 10^(-5)`
Now , `p_(O_(2)) = k_(H) xx x_(O_(2)) = (46.82 xx 10^(3)"bar") xx (3.37 xx 10^(-5))`
` = 1.58` bar .