Question

Q: A given volume of ozonised oxygen ( containing 60% O2 by volume) required 220 sec
to effuse which an equal volume of oxygen took 200 sec under the same conditions. If
density of Oz is 1.6 g/L, find density of O3.​

Answer 1

Answer:This may help you

Explanation:

Let VmL of gas effused.

V/200

V/220= d mixd O2

​⇒d mix

​=1.6×(1.1) 2

=1.936g/L

Let density of ozone is d; In 100 volume ozonised oxygen, 60% O2 and 40% by volume O3 is present.

∴ Mass of mixture=mass of ozone + mass of oxygen

100×1.936=40×d+60×1.6

Density of O3 =2.44g/L.

Answer 2

The answer is the Density of O3 is 2.44g/L

Given:

The volume of ozonised oxygen ( containing 60% O2 by volume) is 220 sec

The density of Oz is 1.6g/L

To Find:

The density of O3

Solution:

Let me of gas be effused.

V/200⁄V/200

$\sqrt{ \frac{do2}{dmix} }$

mix = 1.6×(1.1)^2 = 1.936g/L

Let density of ozone is d; In 100 volumes of ozonised oxygen, 60% O2 and 40% by volume O3 are present.

∴ Mass of mixture=mass of ozone + mass of oxygen

100×1.936 = 40×d+60×1.6

The density of O3 = 2.44g/L.

Hence,

Hence, The Density of O3 is = 2.44g/L.