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 is effused
`(V//220)/(V//200)=sqrt((d_(O_(2)))/(d_("mix")))rArr1.936` g/L
Let density of ozone is d. In 100 volume ozonised oxygen, 60% `O_(2)` and 40% by volume `O_(3)` is present
`therefore` 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