Question

The volume of the 64 gms of oxygen occupied at S.T.P is

Answer 1

Hey Friend,

Atomic weight of oxygen = 16

At STP, 
1 mole of O2 = 16 x 2
                      = 32 g

We know,
1 mole = 22.4 litres
1 mole of O2 = 22.4 litres
32 g of O2 = 22.4 litres
64 g of O2 = 64 x 22.4 / 32 
                  = 44.8 litres

Therefore, from the above calculations, we can infer that 64 g of oxygen is 44.8 litres at STP.

Hope it helps!

Answer 2

Answer:

Atomic mass of O=16

Explanation:

1Mole of O2=16*2=32

we know that

1 mole of O2 =22.4 litres

32gm of O2 =22.4/32 litres

64gm of O2= 22.4/32*64 litres =44.8 litres

Hence , the volume of the 64 grams of oxygen occupied at the STP is 44.8 litres .