Question

$600ml$ of a mixture of ${O_3}$ ​ and ${O_2}$ weighs $1g$ at NTP. Calculate the volume of ozone in the mixture. $105ml$ of pure water $({4^o}C)\;$ is saturated with $N{H_3}$ ​ gas, producing a solution of density $0.9gm{l^{ - 1}}.$

Answer 1

We know that oxygen is a non-metal which is present in the second period and p-block of the modern periodic table. To solve this question, we will first assume the volume of the ${O_2}$ molecule and then we will calculate its weight using its volume. Using the weight of oxygen molecules, we will then find out the weight of ozone molecules present in the mixture.

Complete answer:
We know that oxygen is a non-metal which is present in the second period and p-block of the modern periodic table. When two atoms of oxygen join together to form a molecule, then it is known as an oxygen molecule. When three atoms of oxygen are present in a molecule together, then it is known as an ozone molecule.
Molar mass of an oxygen molecule is $32gmo{l^{ - 1}}.$
Molar mass of an ozone molecule is $48gmo{l^{ - 1}}.$
Now, let us assume the volume of ${O_2}$ in mixture be x ml.
At STP, the weight of x ml ${O_2}$ is: $\dfrac{{32 \times x}}{{22400}}g$
Now, the volume of ${O_3}$ in the mixture will be $(600 - x)ml.$
Weight of the ozone molecule is: $\dfrac{{48 \times (600 - x)}}{{22400}}g$
Now, the sum of these two weights will be equal to one.
$\dfrac{{48 \times (600 - x)}}{{22400}} + \dfrac{{32 \times x}}{{22400}} = 1$
$2x + 28800 - 48x = 22400$
Now, solving this equation for x, we get
$x = 400ml$
Therefore, the volume of ${O_3}$ in the mixture is:
$600 - 400 = 200ml.$

Note:
Ozone is a molecule which is made up of three oxygen atoms. It has a pale blue colour and a pungent smell. It is the protective layer present in the earth’s atmosphere. It protects the earth from the harmful radiations (ultraviolet rays) coming from the sun because UV rays can cause many skin disorders (like skin burn) and many diseases.