Question

how many molecules of oxygen are present in 1 litre air containing 80% volume of 02 at STP

Answer 1

At S.T.P., 1 mole of any gas = 22.4 Litres(It's a given. I assume you know this one.) 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% = 0.0446 x 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 Oxygen gas.

Answer 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.

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.