
Ozone is getting converted into oxygen in the chamber. The Average molecular weight of the mixture is 40. If 10 y% of ozone is decomposed. What is y?
Answer
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Hint: To solve this question firstly write the decomposition reaction of ozone into oxygen which is written as ${{O}_{3}}\to {{O}_{2}}+[O]$. Mole ratio of each one will be 1:1:1 and the molecular weight of oxygen is 32. Molecular weight ozone is 48 g/mol. Apply the stoichiometry rule here.
Complete step by step solution:
Ozone is an unstable molecule which is made up of three oxygen atoms. In this question it is told that decomposition of ozone$({{O}_{3}})$ is taking place so, the decomposition reaction that will take place is,
\[{{O}_{3}}\to {{O}_{2}}+[O]\]
Here we have taken ozone $({{O}_{3}})$ as inlet which decomposed into the mixture of ${{O}_{2}}$ and [O] inside the chamber. In the question the average molecular weight of the mixture is given as 40g. So, we can take the total weight of the mixture as 80g. Now, as the weight of oxygen is not given so, let us take the weight of ${{O}_{2}}$ as x g and therefore the weight of [O] will be (80-x) g. And the weight of ozone which will produce 80 g of mixture is 10 y% that is 0.1 y g. Therefore
\[{{O}_{3}}\to {{O}_{2}}+[O]\]
Conc. 0.1y x (80-x)
And they all are present in 1:1:1 mole ratio.
Here, we can see 32 g of ${{O}_{2}}$ will get produced by 48 g of ${{O}_{3}}$
So, x g of ${{O}_{2}}$ will get produced by = $\dfrac{48}{32}x=1.5x\,g\,{{O}_{3}}$
And, 16 g of [O] will be produced by 48 g of ${{O}_{3}}$,
Thus, (80-x) g of [O] will be produced by = $\dfrac{48}{16}(80-x)=(240-3x)g\,{{O}_{3}}$
Therefore the two equation will be formed here that is,
1.5x = 0.1y………… (1)
240-3x = 0.1y …………….. (2)
Equate the equation, we will get x = 53.33 and y = 800
Thus the Y is 800.
Note: ${{O}_{2}}$ is the most stable form of molecular oxygen and it combines with atomic oxygen which is very reactive to form the unstable molecule that is ozone. Ozone absorbs the UV radiations which have very high energy to break the bond of ozone and resulting to form ${{O}_{2}}$ and one free oxygen molecule. And then this free oxygen again reacts with ${{O}_{2}}$ to form another ozone molecule.
Complete step by step solution:
Ozone is an unstable molecule which is made up of three oxygen atoms. In this question it is told that decomposition of ozone$({{O}_{3}})$ is taking place so, the decomposition reaction that will take place is,
\[{{O}_{3}}\to {{O}_{2}}+[O]\]
Here we have taken ozone $({{O}_{3}})$ as inlet which decomposed into the mixture of ${{O}_{2}}$ and [O] inside the chamber. In the question the average molecular weight of the mixture is given as 40g. So, we can take the total weight of the mixture as 80g. Now, as the weight of oxygen is not given so, let us take the weight of ${{O}_{2}}$ as x g and therefore the weight of [O] will be (80-x) g. And the weight of ozone which will produce 80 g of mixture is 10 y% that is 0.1 y g. Therefore
\[{{O}_{3}}\to {{O}_{2}}+[O]\]
Conc. 0.1y x (80-x)
And they all are present in 1:1:1 mole ratio.
Here, we can see 32 g of ${{O}_{2}}$ will get produced by 48 g of ${{O}_{3}}$
So, x g of ${{O}_{2}}$ will get produced by = $\dfrac{48}{32}x=1.5x\,g\,{{O}_{3}}$
And, 16 g of [O] will be produced by 48 g of ${{O}_{3}}$,
Thus, (80-x) g of [O] will be produced by = $\dfrac{48}{16}(80-x)=(240-3x)g\,{{O}_{3}}$
Therefore the two equation will be formed here that is,
1.5x = 0.1y………… (1)
240-3x = 0.1y …………….. (2)
Equate the equation, we will get x = 53.33 and y = 800
Thus the Y is 800.
Note: ${{O}_{2}}$ is the most stable form of molecular oxygen and it combines with atomic oxygen which is very reactive to form the unstable molecule that is ozone. Ozone absorbs the UV radiations which have very high energy to break the bond of ozone and resulting to form ${{O}_{2}}$ and one free oxygen molecule. And then this free oxygen again reacts with ${{O}_{2}}$ to form another ozone molecule.
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