How many molecules of O2 are present in 1 L of air containing 80% volume of O2 at STP??
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ita air to it will be molecules of O2. At S.T.P., 1 mole of any gas = 22.4 Litres(It's a given. I assume you know thisone.) Therefore, 1 Litre of a gas = 1/22.4 = 0.0446 moles Out of these 0.0446 moles, 80% is Oxygen gas. Number of moles of Oxygen gas = 0.0446 x 80% = 0.0446x 80/100 = 0.0357 moles. 1 mole of any gas = 6.02 x 1023 units (This is also a given, known as the Avagadro's constant.) Therefore, 0.0357 moles = 0.0357 x 6.02 x 1023 = 2.15 x 1022 molecules of O2 is present in 1 litre air
ita air to it will be molecules of O2. At S.T.P., 1 mole of any gas = 22.4 Litres(It's a given. I assume you know thisone.) Therefore, 1 Litre of a gas = 1/22.4 = 0.0446 moles Out of these 0.0446 moles, 80% is Oxygen gas. Number of moles of Oxygen gas = 0.0446 x 80% = 0.0446x 80/100 = 0.0357 moles. 1 mole of any gas = 6.02 x 1023 units (This is also a given, known as the Avagadro's constant.) Therefore, 0.0357 moles = 0.0357 x 6.02 x 1023 = 2.15 x 1022 molecules of O2 is present in 1 litre air
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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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