Pure `O_(2)` diffuses through an aperture in 224 second, whereas mixture of `O_(2)` and another gas containing 80% `O_(2)` diffuses from the same in 234 second. The molecular mass of gas will be:

Explanation

The gaseous mixture contains 80% `O_(2)` and 20% gas.
`:.` Average molecular weight of mixture.
`(M_("mix")) = (32 xx 80 + 20 xxm_("mix"))/( 100) ` ......(i)
Now for diffusion of gaseous mixture and pure `O_(2)`
`(r_(O_(2)))/( r_(m)) = sqrt((M_("mix"))/(M_(O_(2))))` or `(V_(O_(2)))/( V_("mix")) xx( t_("mix"))/( t_(O_(2)))= sqrt((M_("mix"))/( 32))` or `(1)/( 224) xx( 234)/( 1) sqrt((M_("mix"))/( 32)) ` ....(ii)
`:. M_("mix") = 34.92`
By (i) and (ii) mol weight of gas ( m ) = 46.6