The density of a gaseous mixture `O_(2)` and `N_(2)` is 1.4 g/L at S.T.P. Find out the partial pressure of `O_(2)` in the mixture.

Explanation

Gram molecular weight of the mixture = d at S.T.P. (in g/L) `xx22.4 = 1.4 xx 22.4 = 31.36` g
Molecular weight of a mixture ` = M_(A).x_(A) + M_(B).x_(B)`
Substituting he given data,
`31.36 = 32X_(O_(2)) + 28(1-X_(O_(2)))`
Mole fraction of oxygen, `X_(O_(2)) = 0.84`
Partial pressure of oxygen, `P_(O_(2)) = X_(O_(2)) . P = 2.84 xx 1 = 0.84 ` atm