A sample of online oxygen contains 60% ozone by weight. What is the average molecular mass (amu) of the sample? (Atomic mass of O=16)
Answers
Answered by
0
Answer:
Let VmL of gas effused.
V/200
V/220
=
d
mix
d
O
2
⇒d
mix
=1.6×(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
100×1.936=40×d+60×1.6
Density of O
3
=2.44g/L
Explanation:
Similar questions