Question

600 ml, of a mixture of O3 and O2 weights 1 g at NTP.Calculate the volume of Ozone in the mixture.

Answer 1

Explanation:

Let us consider the volume of O

2

=x ml

At STP, weight of x ml O

2

=

22400

32×x

gm

Weight of (600−x) ml of O

3

=

22400

48×(600−x)

gm

∴

22400

32x

+

22400

48×(600−x)

=1

⇒32x+28800−48x=22400

⇒x=400 mL

∴ Hence the volume of O

3

• =600−400=200 ml