A given volume of ozonised oxygen (containing 60% oxygen by volume ) required 220 sec to effuse while an equal volume of oxygen took 200 sec only under identical conditions. If density of `O_(2)` is 1.6 g/L then find density of `O_(3)`.

Options

A. `1.936 g//L`
B. `2.16 g//L`
C. ` 3.28 g//L`
D. ` 2.24 g//L`

Answer

D

Explanation

Let V mL of gas effused
`(V//220)/( V//200) = sqrt(( d_(O_(2)))/( d_("mix"))) rArr d_("mix) 1.6 xx ( 1.1)^(2) =1.936g//L`
Let density of ozone is d, In 100 volume ozonised oxygen, 60% `O_(2)` and 40% by volume `O_(3)` is present
`:.` Mass of mixture = mass of ozone `+ ` mass of oxygen
`100 xx 1.936 = 40 xx d + 60 xx 1.6`
density of `O_(3)` is `2.44 g//L`