Question

Six hundred milliliters of ozonized oxygen at STP was found to weigh $1g$. What is the volume of ozone in the ozonized oxygen?
(A) $200ml$
(B) $150ml$
(C) $100ml$
(D) $50ml$

Answer 1

Hint: We know that ozonized oxygen contains ozone and oxygen in it. To solve this problem, first assume the volume of ozone present. To find the volume of oxygen, subtract this volume from the total volume of ozonized oxygen. While solving the problem, remember that a mole of any gas occupies $22.4L$ at standard temperature and pressure.

Complete answer:
Ozone is formed when a silent electric discharge is passed through dry oxygen. Oxygen can never be converted into ozone completely. We always get a mixture of oxygen and ozone called ozonized oxygen.
Let the volume of ozone in ozonated oxygen be $x{\text{ }}ml$
So, the volume of ${O_2}$ present in ozonized oxygen $ = (600 - x)ml$
We know that a mole of any gas occupies $22.4L$ at standard temperature and pressure.
One mole of ${O_3} = 48{\text{ }}g$
One mole of ${O_2} = 32{\text{ }}g$
So, $22400ml$ of ${O_3}$ and ${O_2}$ at STP will weigh $48$ and $32{\text{ }}g$ respectively.
The weight of $x{\text{ }}ml$ of ${O_3} = \dfrac{{x \times 48}}{{22400}}g$
The weight of $(600 - x)ml$ of ${O_2} = \dfrac{{\left( {600 - x} \right)}}{{22400}} \times 32$
The total weight of ozonized oxygen is:
$\dfrac{{48x}}{{22400}} + \dfrac{{\left( {600 - x} \right) \times 32}}{{22400}} = 1.0$
$x = 200{\text{ }}ml$
The volume of ozone in the ozonized oxygen is $200ml$
Therefore, option A is the correct answer.

Note:
Remember that ozonized oxygen is a mixture of oxygen and ozone. To solve these types of problems, first assume the volume of one component of the mixture. Then to find the volume of the other component of the mixture, subtract the assumed volume from the total volume of the mixture.