How many molecules of O2 are present in 1L air containing 80% volume at STP
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Answered by
11
80% of the volume of air contains oxygen
so volume of oxygen = 0.8 L
at STP 22.4 L of oxygen contains 6.023×10²³ molecules of O₂
so at STP 0.8 L of oxygen contains (6.023×10²³/22.4)=2.69 ×10²³ molecules of O₂
so volume of oxygen = 0.8 L
at STP 22.4 L of oxygen contains 6.023×10²³ molecules of O₂
so at STP 0.8 L of oxygen contains (6.023×10²³/22.4)=2.69 ×10²³ molecules of O₂
Answered by
2
Answer:
Answer: 2.15x 10^22 molecules.
Explanation:
▪️ Correct question should be: how many molecules of oxygen are present in 1 litre of air containing 80% of O2.
First, we will have to find amount of energy in one oxygen.
Therefore,
➡️ Amount of oxygen present in 1 litre air is = 80% *1000 ml
➡️ (80/100)*1000
=800ml
▪️ We know that, 1 mole of STp occupies 22.4 litre
It is equal to = 22400ml of O2 gas contains 6.022x 10^23 molecules
So,
➡️ 1ml of O2 gas Contains =6.022x10^23/22400
According to question, we have to find 800ml molecules.
Therefore,
800ml contains =(6.022x 10^23/22400) *800
=> 2.15 x 10^22
▪️ Hence, molecules present in 1 litre air containing 80% volume of at STP is 2.15x 10^22 molecules.
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