Question

what is the volume occupied by the following gases at S.T.P?
64g of oxygen​

Answer 1

Answer:

44.8 L is the volume occupied by the following gases at S.T.P.

Explanation:

One mole of oxygen gas occupies a volume at STP = 22.4 L

Now,

By using the formula of mole, we find how much moles of oxygen are present,

Mole = Given mass / molar mass

The number of moles in 64 g of oxygen gas = Given mass / molar mass

=> Mole = 64 / 32

=> Mole = 2

.°. There is 2 moles of oxygen in 64 g.

Therefore, the volume occupied by 2 moles of oxygen gas

=> 2 x 22.4 L = 44.8 L