A given volume of ozonized oxygen (containing 60% oxygen by volume) required 220 sec to effuse which an equal volume of oxygen took 200 sec only under the 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. 2.44 g/L

Answer

D

Explanation

Let V mL of gas effused , `(V//220)/(V//200) = sqrt((d_(O_2))/(d_("mix"))) implies d_("mix") = 1.6 xx (1.1)^(2) = 1.936` g/L
Let density of ozone is d , ln 100 volume ozonized oxygen , 60% and 40% by volume `O_(2)` is presence
`therefore` Mass of mixture = mass of ozone + mass of oxygen
`100 xx 1.936 = 40 xx d + 60 xx 1.6 implies` density of `O_(3)` is 2.44 g/L