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)`.
A
1.963 g/L
B
2.16 g/L
C
3.28 g/L
D
2.24 g/L
Text Solution
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The correct Answer is:
d
(d) Let V mL of gas effused `(V//220)/(V//220)=sqrt((d_(O_(2)))/(d_("mix")))impliesd_("mix")1.6xx(1.1)^(2)=1.936g//L` Let density of ozome 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 `100xx1.936=40xxd+60xx1.6` density of `O_(3)` is 2.44 g/L
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